Question 12 Marks
Find:
$\int\frac{\text{dx}}{5 - \text{8x - x}^{2}}$
Answer$\int\frac{\text{dx}}{5 - \text{8x - x}^{2}} = \int\frac{\text{dx}}{(\sqrt{21)^{2} - (\text{x + 4)}^{2}}}$
$= \frac{1}{2\sqrt{21}} \log \bigg|\frac{\sqrt{21} + \text{(x + 4)}}{\sqrt{21} - \text{(x + 4)}}\bigg| + \text{c}$
View full question & answer→Question 22 Marks
Find:
$\int \frac{\text{dx}}{\sqrt{3 - \text{2x - x}^{2}}}$
Answer$\text{I} = \int \frac{\text{dx}}{\sqrt{(2)^{2} - \text{(x + 1)}^{2}}}$
$= \sin^{-1} \bigg(\frac{\text{x + 1}}{2}\bigg) + \text{c}$
View full question & answer→Question 32 Marks
Find $\int \frac{\text{d}x}{x^{2} + 4x + 8}$
Answer$\int \frac{\text{dx}}{\text{x}^{2} + \text{4x + 8}} = \int \frac{\text{dx}}{(\text {x + 2)}^{2} + (2)^{2}}$
$= \frac{1}{2} \tan^{-1} \frac{\text{x + 2}}{2} + \text{C}$
View full question & answer→Question 42 Marks
Evaluate: $\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\cos^2\text{x}}\text{dx}$
Answer$\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\cos^2\text{x}}$
$\int\frac{2\cos^2\text{x}-1+2(1-\cos^2\text{x})}{\cos^2\text{x}}\text{dx}$
$\int\Big(2-\frac{1}{\cos^2\text{x}}+\frac{2}{\cos^2\text{x}}-1\Big)\text{dx}$
$\int\Big(1+\frac{1}{\cos^2\text{x}}\Big)\text{dx}$
$\text{x}+\int\text{sec}^2\text{x}\text{dx}$
$\text{x}+\tan\text{x}+\text{C}$
View full question & answer→Question 52 Marks
Find: $\int\sin\text{x}.\log\cos\text{x}\text{dx}.$
Answer$\int\sin\text{x}.\log\cos\text{x}\text{dx}=$
$\log\cos\text{x}\int\sin\text{x}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}(\log\cos\text{x})\int\sin\text{x}\text{dx}\Big)\text{dx}$
$=-\cos\text{x}\log\cos\text{x}-\int\Big(\frac{\sin\text{x}}{\cos\text{x}}\times\cos\text{x}\Big)\text{dx}=-\cos\text{x}\log\cos\text{x}+\cos\text{x}+\text{C}.$
View full question & answer→Question 62 Marks
Find: $\int\frac{\tan^2\text{x}\sec^2\text{x}}{1-\tan^6\text{x}}\text{dx}.$
AnswerLet $\text{I}=\int\frac{\tan^2\text{x}\sec^2\text{x}}{1-\tan^6\text{x}}\text{dx}$
Let $\tan^3\text{x}=\text{t}$
$3\tan^2\text{x}\sec^2\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{1}{3}\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{3}\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{6}\text{ln}\Big|\frac{1+\text{t}}{1-\text{t}}\Big|+\text{c},$
$\text{I}=\frac{1}{6}\text{ln}\Big|\frac{1+\tan^3\text{x}}{1-\tan^3\text{x}}\Big|+\text{c}$
View full question & answer→Question 72 Marks
Find: $\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}.$
Answer$\text{I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
So, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}$
Or, $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$
Since, we know
$\int\frac{\text{dx}}{\sqrt{\text{x}^2+\text{a}^2}}=\text{ln}\ \Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}$
$\text{I}=\text{ln}\ \Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
i.e., $\text{I}=\text{ln}|\tan\text{x}+\sqrt{{\tan}^2\text{x}+4}|+\text{C}$
View full question & answer→Question 82 Marks
Find: $\int\sin^{-1}(2\text{x})\text{dx}.$
Answer$\int\sin^{-1}(2\text{x})\text{dx}$
Using ILATE rule
$\text{x}\sin^{-1}2\text{x}-\int\frac{2\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{-8\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
Taking $1-4\text{x}^2=\text{t}$
$\Rightarrow-8\text{x}\times\text{dx}=\text{dt}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\frac{2\text{t}^{\frac{1}{2}}}{1}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\text{t}^{\frac{1}{2}}}{2}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\sqrt{1-4\text{x}^2}}{2}+\text{C}$
View full question & answer→Question 92 Marks
Integrate the function in Exercise:$\cos^{3}\text{x}\ \text{e}^{\log\sin\text{x}}$
Answer$\cos^{3}\ \text{xe}^{\log\sin\text{x}}=\cos^{3}\text{x}\times\sin\text{x}$
$\text{Let}\ \cos\text{x}=\text{t}\Rightarrow-\sin\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\int\cos^{3}\text{xe}^{\log\sin\text{x}}\text{dx}=\int\cos^{3}\text{x}\sin\ \text{xdx}$
$=-\int\text{t}^3\ \text{dt}$
$=-\frac{\text{t}^{4}}{4}+\text{C}$
$=-\frac{\cos^{4}\text{x}}{4}+\text{C}$
View full question & answer→Question 102 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^5}\text{dx}$
Answer$\int\text{x}^{-5}\text{dx}$
$=\frac{\text{x}^{-5+1}}{-5+1}+\text{c}$
$=-\frac{1}{4}\text{x}^{-4}+\text{c}$
$=-\frac{1}{4\text{x}^4}+\text{c}$
View full question & answer→Question 112 Marks
Evaluate:
$\int\limits^3_23^\text{x}\text{ dx}$
Answer$\text{I}=\int\limits^3_23^\text{x}$
$=\Big[\frac{3\text{x}}{\log3}\Big]^3_2+\text{C}$ $\Big(\text{Use}:\int\text{a}^{\text{x}}=\frac{\text{a}^{\text{x}}}{\log\text{a}}+\text{C}\Big)$
$=\frac{3^3}{\log3}-\frac{3^2}{\log3}+\text{C}$
$=\frac{1}{\log3}(3^2-3^3)+\text{C}$
$=\frac{1}{\log3}(27-9)+\text{C}$
$=\frac{1}{\log3}(18)+\text{C}$
View full question & answer→Question 122 Marks
Evaluate:
$\int\frac{\text{e}^{6\log_\text{e}\text{x}}-\text{e}^{5\log_\text{e}\text{x}}}{\text{e}^{4\log_\text{e}\text{x}}-\text{e}^{3\log_\text{e}\text{x}}}\text{dx}$
Answer$\int\frac{\text{e}^{6\log_\text{e}\text{x}}-\text{e}^{5\log_\text{e}\text{x}}}{\text{e}^{4\log_\text{e}\text{x}}-\text{e}^{3\log_\text{e}\text{x}}}\text{dx}=\int\frac{\text{x}^6-\text{x}^5}{\text{x}^4-\text{x}^3}\text{dx}$
$=\int\frac{\text{x}^5(\text{x}-1)}{\text{x}^3(\text{x}-1)}\text{dx}$
$=\int\text{x}^2\text{dx}$
$=\frac{\text{x}^3}{3}+\text{C}$
View full question & answer→Question 132 Marks
Evaluate the following integrals:
$\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
Answer$\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
View full question & answer→Question 142 Marks
Integrate the function in exercise.
$\text{x} \ \sin3\text{x}$
AnswerLet $\text{I}=\int\text{x}\sin3\text{x dx}$
Taking x as first function and x as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sin3\text{x dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\text{x}\Big)\int\sin3\text{x dx}\Bigg\}$
$=\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1.\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{3}\int\cos3\text{x} \ \text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{9}\sin3\text{x} \ \text{dx}+\text{C}$
View full question & answer→Question 152 Marks
Evaluate:
$\int\Big(\frac{2\cos^2\text{x}-\cos2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
Answer$\int\Big(\frac{2\cos^2\text{x}-\cos2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{2\cos^2\text{x}-(2\cos^2\text{x}-1)}{\cos^2\text{x}}\Big)\text{dx}$ $[\because\cos2\text{x}=2\cos^2\text{x}-1]$
$=\int\text{sec}^2\text{x}\text{ dx}$
$=\tan\text{x}+\text{c}$
View full question & answer→Question 162 Marks
Prove the following Exercise:
$\int^{1}_{0}\text{x e}^{\text{x}}\ \text{dx}=1$
Answer$\text{Let I}=\int^{1}_{0}\text{x e}^{\text{x}}\ \text{dx}$
Integrating by parts, we obtain
$\text{I}=\text{x}\int^{1}_{0}\text{e}^{\text{x}}\ \text{dx}-\left\{\bigg(\frac{\text{d}}{\text{dx}}\text{(x)}\bigg)\int\text{e}^{\text{x}}\text{dx}\right\}\text{dx}$
$=\Big[\text{xe}^{\text{x}}\Big]^{1}_{0}-\int^{1}_{0}\text{e}^{\text{x}}\ \text{dx}$
$=\Big[\text{xe}^{\text{x}}\Big]^{1}_{0}-\Big[\text{e}^{\text{x}}\Big]^{1}_{0}$
$=\text{e}-\text{e}+1$
$=1$
Hence, the given result is Proved.
View full question & answer→Question 172 Marks
Evaluate the following integrals:
$\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
We know that,
$[\text{x}]=1,\text{ when }1<\text{x}<2$
$\therefore\ \text{I}=\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
$\text{I}=\int\limits^{2}_1(0)\text{dx}$
$\text{I}=0$
View full question & answer→Question 182 Marks
Evaluate the following:
$\int\frac{\text{dx}}{\sqrt{16-9\text{x}^2}}$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\sqrt{16-9\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{(4)^2-(3\text{x})^2}}\text{dx}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\text{x}}{4}\Big)+\text{C}$ $\bigg[\because\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\bigg]$
View full question & answer→Question 192 Marks
If $\int\limits^1_0(3\text{x}^2+2\text{x}+\text{k})\text{dx}=0,$ find the value of k.
AnswerWe have,
$\int\limits^1_0(3\text{x}^2+2\text{x}+\text{k})\text{dx}=0$
$\Rightarrow\big[\text{x}^3+\text{x}^2+\text{kx}\big]^1_0=0$
$\Rightarrow1+1+\text{k}-0=0$
$\Rightarrow\text{k}=-2$
View full question & answer→Question 202 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{4}_0\tan\text{x dx}$
Answer$\int\limits^\frac{\pi}{4}_0\tan\text{x dx}$
$=\big[\log\sec\text{x}\big]^{\frac{\pi}{4}}_0$
$=\log\sec\frac{\pi}{4}-\log\sec0$
$=\log\sqrt{2}-\log1$
$=\log2^{\frac{1}{2}}-0$
$=\frac{1}{2}\log2$
View full question & answer→Question 212 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\cos2\text{x}}\text{ dx}$
Answer$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\cos2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{2\sin^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{2}\sin\text{x}\text{ dx}$
$=-\sqrt{2}\Big[\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$=-\big(0-\sqrt{2}\big)$
$=\sqrt{2}$
View full question & answer→Question 222 Marks
$\int\limits^{1}_0\{\text{x}\}\text{dx},$ where {x} denotes the fractional part of x.
AnswerWe have,
$\text{I}=\int\limits^{1}_0\{\text{x}\}\text{dx}$
We know $\{\text{x}\}=\text{x},0<\text{x}<1$
$\therefore\ \text{I}=\int\limits^{1}_0\{\text{x}\}\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\Big]^1_0$
$=\frac{1}{2}-\frac{0}{2}$
$=\frac{1}{2}$
View full question & answer→Question 232 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{-1}^{1}\text{(x}+1)\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{-1}^{1}\text{(x+1)}\text{dx} $ $\int\text{(x}+1)\text{dx}=\frac{\text{x}^{2}}{2}+\text{x}$ By second theorem of calculus, we obtain $\text{I}=\text{F}(1)-\text{F}(-1)$$=\bigg(\frac{1}{2}+1\bigg)-\bigg(\frac{1}{2}-1\bigg)$
$=\frac{1}{2}+1-\frac{1}{2}+1$
$=2$
View full question & answer→Question 242 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\cos\text{x}}\text{ dx}$
AnswerWe use $1+\cos\text{x}=2\cos^2\frac{\text{x}}{2}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{2}\cos\frac{\text{x}}{2}\text{ dx}$
$=\sqrt{2}\Big[2\sin\frac{\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\sqrt{2}\Big[\frac{1}{\sqrt{2}}\Big]$
$=2$
View full question & answer→Question 252 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
AnswerWe have
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
We know that $\int\text{e}^{-\text{x}}=-\text{e}^{-\text{x}}$
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
$=\big[-\text{e}^{-\text{x}}\big]^{\infty}_0$
$=\big[\text{e}^{-\infty}+\text{e}^{-0}\big]$ $[\because\text{e}^{\infty}=0,\text{ e}^0=1\big]$
$=[-0+1]$
$=1$
View full question & answer→Question 262 Marks
Evaluate the following integrals:
$\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^1_0\frac{2\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $1+\text{x}^2=\text{t}$
$2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow1$
And $\text{x}\rightarrow1;\text{ t}\rightarrow2$
$\therefore\ \text{I}=\int\limits^1_0\frac{\text{dt}}{\text{t}^2}$
$=\Big[\log_{\text{e}}|\text{t}|\Big]^2_1$
$=\log_{\text{e}}2-\log_{\text{e}}1$
$=\log_{\text{e}}2-0$
$=\log_{\text{e}}2$
View full question & answer→Question 272 Marks
Write a value of $\int\cos^4\text{x }\sin\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\cos^4\text{x }\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\text{I}=-\int\text{t}^{4}\text{ dt}$
$=-\frac{\text{t}^5}{5}+\text{C}$
$\text{I}=\frac{\cos^5\text{x}}{5}+\text{C}$
View full question & answer→Question 282 Marks
$\int\cos^2\frac{\text{x}}{2}\text{dx}$
Answer$\int\cos^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\frac{1+\cos\text{x}}{2}\Big)\text{dx}$ $\Big[\therefore\cos^2\frac{\text{x}}{2}=\frac{1+\cos\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1+\cos\text{x})\text{dx}$
$=\frac{1}{2}[\text{x}+\sin\text{x}]+\text{C}$
View full question & answer→Question 292 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{2}^{3}\frac{\text{x}\ \text{dx}}{\text{x}^{2}+1}$
Answer$\text{Let}\text{I}=\int\limits_{2}^{3}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
$\int\frac{\text{x}}{\text{x}^{2}+1}\text{dx}=\frac{1}{2}\int\frac{2\text{x}}{\text{x}^{2}+1}\text{dx}=\frac{1}{2}\text{log}(1+\text{x}^{2})=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(3)-\text{F}(2)$
$=\frac{1}{2}\big[\text{log}\big(1+(3)^{2}\big)-\text{log}\big(1+(2)^{2}\big)\big]$
$=\frac{1}{2}\big[\text{log}(10)-\text{log(5)}\big]$
$=\frac{1}{2}\text{log}\bigg(\frac{10}{5}\bigg)=\frac{1}{2}\text{log}2$
View full question & answer→Question 302 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$
$=\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}\ \ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\text{x}\ \text{dx}$
Adding eq.(i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}(\cos^{2}\text{x}+\sin^{2}\text{x})\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)$
$\Rightarrow\ 21=\frac{\pi}{2}\ \ \ \Rightarrow\ \ \text{I}=\frac{\pi}{4}$
View full question & answer→Question 312 Marks
Integrate the function in Exercise:
$\frac{\cos\text{x}}{\sqrt{4-\sin^{2}}\text{x}}$
Answer$\frac{\cos\text{x}}{\sqrt{4-\sin^{2}\text{x}}}$$\text{Let}\ \sin\text{x}=\text{t}\Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}}{\sqrt{4-\sin^{2}\text{x}}}\text{dx}=\int\frac{\text{dt}}{\sqrt{(2)^{2}-\text{(t)}^{2}}}$
$=\sin^{-1}\bigg(\frac{\text{t}}{2}\bigg)+\text{C}$
$=\sin^{-1}\bigg(\frac{\sin\text{X}}{2}\bigg)+\text{C}$
View full question & answer→Question 322 Marks
Evaluate the following integrals:
$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$
Answer$\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$ Let $\tan^{-1}\text{x}=\text{t}$ $\Rightarrow\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$ Now, $\int\frac{\sin(\tan^{-1}\text{x})}{1+\text{x}^2}\text{ dx}$$=\int\sin\text{t dt}$
$=-\cos(\text{t})+\text{C}$
$=-\cos\big(\tan^{-1}\text{x}\big)+\text{C}$
View full question & answer→Question 332 Marks
If $\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt},$ the write the value of f'(x).
Answer$\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt}$
$\Rightarrow\text{f(x)}=\text{t}\big[-\cos\text{t}\big]^{\text{x}}_0-\int\limits^{\text{x}}_0\frac{\text{d}}{\text{dt}}(\text{t})\times(-\cos\text{t})\text{dt}$
$\Rightarrow\text{f(x)}=-(\text{x}\cos\text{x}-0)+\int\limits^{\text{x}}_0\cos\text{t dt}$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\big[\sin\text{t}\big]^{\text{x}}_0$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+(\sin\text{x}-0)$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\sin\text{x}$
Differentiating both sides with respect to x, we get
$\text{f}'(\text{x})=-\big[\text{x}\times(-\sin\text{x})+\cos\text{x}\times1\big]+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-(-\text{x}\sin\text{x})-\cos\text{x}+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\text{x}\sin\text{x}$
Thus, the value of f'(x) is x $\sin\text{x}$
View full question & answer→Question 342 Marks
Evaluate the following integrals:
$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}$
Answer$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}=\int^\limits{\pi}_{0}\sin\text{x }\text{dx}+\int^\limits{2\pi}_{\pi}-\sin\text{x }\text{dx}$
$=\big[-\cos\text{x}\big]^{\pi}_0+\big[\cos\text{x}\big]^{2\pi}_\pi$
$=\big[1+1\big]+\big[1+1\big]$
$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}=4$
View full question & answer→Question 352 Marks
Evaluate the following integrals:
$\int\sin^5\text{x}\cos\text{x dx}$
Answer$\int\sin^5\text{x}\cos\text{x dx}$
$\text{Let }\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\text{Now,}\int\sin^5\text{x}\cos\text{x dx}$
$=\int\text{t}^5\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{1}{6}\sin^6\text{x}+\text{C}$
View full question & answer→Question 362 Marks
Evaluate the following integrals:
$\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}\big]\text{dx}+\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{1.5}_0(1)\text{dx}$ $\begin{bmatrix}\because\big[\text{x}\big]=\begin{cases}0,&0\leq\text{x}<1\\1,&1\leq\text{x}<1.5\end{cases}\end{bmatrix}$
$=0+\big[\text{x}\big]^{1.5}_1$
$=1.5-1$
$=\frac{1}{2}$
View full question & answer→Question 372 Marks
Integrate the functions in Exercises:
$\frac{1}{\sqrt{9-25\text{x}^2}}$
Answer$\int\frac{1}{\sqrt{9-25\text{x}^2}}\text{ dx}=\int\frac{1}{\sqrt{(3)^2-(5\text{x})^2}}\text{ dx}$
$=\frac{\sin^{-1}\frac{5\text{x}}{3}}{5\rightarrow\text{Coeff. of x}}+\text{c}$$\ \ \ \ \ \ \ \bigg[\because\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\frac{1}{5}\sin^{-1}\bigg(\frac{5\text{x}}{3}\bigg)+\text{c}$
View full question & answer→Question 382 Marks
Integrate the function in exercise.
$\text{x} \ \sin\text{x}$
AnswerLet $\text{I}=\int\text{x}\sin\text{x dx}$
Taking x as first function and x as second function and integrating by parts, we obtain.
$\text{I}=\int\text{x}\sin\text{x dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\text{x}\Big)\int\sin\text{x dx}\Bigg\}\text{dx}$
$=\text{x}(-\cos\text{x})-\int1.(-\cos\text{x})\text{dx}$
$=-\text{x}\cos\text{x}+\sin\text{x}+\text{C}$
View full question & answer→Question 392 Marks
Evaluate the following integrals:
$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
Answer$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$\text{Let }1+\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\text{Now,}\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$=\int\sqrt{\text{t}}\text{ dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{2}{3}(1+\text{e}^\text{x})^\frac{3}{2}+\text{C}$
View full question & answer→Question 402 Marks
Integrate the function in Exercise:
$\text{e}^\text{x}\Bigg(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Bigg)$
Answer$\text{I}=\int\text{e}^\text{x}\Bigg[\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Bigg]\text{dx}$
Also, let $\frac{1}{\text{x}}=\text{f}(\text{x})\Rightarrow \ \text{f}'(\text{x})=\frac{-1}{\text{x}^2}$
It is known that, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^\text{x}}{\text{x}}+\text{C}$
View full question & answer→Question 412 Marks
Evaluate the following integrals:
$\int\bigg (2^{\text{x}}+\frac{5}{\text{x}}-\frac{1}{\text{x}^{\frac{1}{3}}}\bigg)\text{dx}$
Answer$\int\bigg(2^{\text{x}}+\frac{5}{\text{x}}-\frac{1}{\text{x}^{\frac{1}{3}}}\bigg)\text{dx}$
$=\int2^{\text{x}}\text{dx}+5\int\frac{1}{\text{x}}\text{dx}-\int\frac{1}{\text{x}^{\frac{1}{3}}}\text{dx}$
$=\frac{2^{\text{x}}}{\log2}+5\log\text{x}-\frac{3}{2}\text{x}^{\frac{2}{3}}+\text{C}$
View full question & answer→Question 422 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_0\cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_0\cos^2\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1+\cos2\text{x}}{2}\text{dx}$
$= \frac{1}{2}\int\limits^\frac{\pi}{2}_0(1+\cos2\text{x})\text{dx}$
$= \frac{1}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{2}\Big[\frac{\pi}{2}+0\Big]$
$=\frac{\pi}{4}$
View full question & answer→Question 432 Marks
Write tha value of $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$
Answer$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$
View full question & answer→Question 442 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{4}^{5}\text{e}^{\text{x}}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{4}^{5}\text{e}^{\text{x}}\ \text{dx}$ $\int\text{e}^\text{x}\ \text{dx}=\text{e}^\text{x}=\text{F}\text{(x)}$By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(5)-\text{F}(4)$
$=\text{e}^{5}-\text{e}^{4}$
$=\text{e}^{4}(\text{e}-1)$
View full question & answer→Question 452 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$ Then,
$\text{I}=\big[-\cos\text{x}+\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+1-(-1+0)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 462 Marks
Write a value of $\int\tan\text{x}\sec^3\text{x dx}$
AnswerLet $\text{I}=\int\tan\text{x}\sec^3\text{x dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x dx}=\text{dt}$
$\text{dx}=\frac{\text{dt}}{\sec\text{x}\tan\text{x}}$
$\therefore\ \text{I}=\int\sec^2\text{x}\tan\text{x dx}$
$=\int\text{t}^2\text{ dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{\sec^3\text{x}}{3}+\text{C}$
$\therefore\ \text{I}=\frac{\sec^3\text{x}}{3}+\text{C}$
View full question & answer→Question 472 Marks
Integrate the function in Exercise.
$\sqrt{4-\text{x}^2}$
Answer$\int\sqrt{4-\text{x}^2}\text{dx}=\int\sqrt{2^2-\text{x}^2}\text{dx}$
$=\frac{\text{x}}{2}\sqrt{2^2-\text{x}^2}+\frac{2^2}{2}\sin^{-1}\frac{\text{x}}{2}+\text{C}$
$\Bigg[\therefore\ \sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Bigg]$
$=\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 482 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$\Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2} \ \&\ 1+\cos\text{x}=2\cos^2\frac{\text{x}}{2}\Big]$
$=\int\tan\frac{\text{x}}{2}\text{dx}$
$=-2\text{ln}\Big|\cos\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 492 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
Answer$\text{Let}\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)-\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}{1+\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos\text{x}-\sin\text{x}}{1+\cos\text{x}\sin\text{x}}\text{dx}=-\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
Adding eq. (i) and (ii), we have $21=0 \ \ \Rightarrow 1=0$
View full question & answer→Question 502 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sin^{2}\text{x dx}$
Answer$\int\limits^\frac{\pi}{2}_0\sin^2\text{x}\text{dx}$
$= \int\limits^\frac{\pi}{2}_0\frac{1-\cos2\text{x}}{2}\text{dx}$
$= \frac{1}{2}\int\limits^\frac{\pi}{2}_0(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{2}(\frac{\pi}{2}-0)$
$=\frac{\pi}{4}$
View full question & answer→Question 512 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{xf}\big(\text{x}^2\big)\text{dx}=0$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{xf}(\text{x}^2)\text{dx}$
Let $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=(-\text{x})\text{f}(-\text{x})^2=-(\text{x})\text{f}(\text{x})^2=-\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=0$ $\bigg[\text{Using}\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=0\text{ when g(x) is odd}\bigg]$
View full question & answer→Question 522 Marks
Integrate the function in Exercise:$\text{e}^{3}\log\text{x}\big(\text{x}^{4}+1\big)^{-1}$
Answer$ \text{e}^{3\log\text{x}}.\big(\text{x}^{4}+1\big)^{-1}=\text{e}^{\log\text{x}^{3}}.\big(\text{x}^{4}+1\big)^{-1}=\frac{\text{x}^{3}}{\big(\text{x}^{4}+1\big)}$
$\text{Let}\ \text{x}^{4}+1=\text{t}\Rightarrow 4\text{x}^3\ \text{dx}=\text{dt}$
$\Rightarrow\int\text{e}^{3\log\text{x}}\big(\text{x}^{4}+1\big)^{-1}\text{dx}=\int\frac{\text{x}^{3}}{\big(\text{x}^{4}+1\big)}\text{dx}$
$=\frac{1}{4}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{4}\log|\text{t}|+\text{C}$
$=\frac{1}{4}\log\big(\text{x}^{4}+1\big)+\text{C}$
View full question & answer→Question 532 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}$
$\text{Here}\ \ \text{f}\text{(x)}=\sin^{7}\text{x}$
$\therefore\ \ \text{f}(-\text{x)}=\sin^{7}(-\text{x})=(-\sin^{7}\text{x})=-\sin^{7}\text{x}=-\text{f}\text{(x)}$
$\therefore\ \ \text{f}(\text{x})\ \text{is an odd function of x.}$
$\therefore\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}=0\ \ \bigg[\because\int^{\text{a}}\limits_{-\text{a}}\text{f}\text{(x)}\text{dx}=0,$when $\text{f(x)}$ is an odd function$\bigg]$
View full question & answer→Question 542 Marks
Evaluate $\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=2\int\sin\text{t dt}$
$=-2\cos\text{t}+\text{C}$
$\text{I}=-2\cos\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 552 Marks
Evaluate $\int\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta$
$\therefore\ \text{I}=\int\frac{\cos\theta}{\cos\theta}\text{ d}\theta$
$=\int\text{d}\theta$
$=\theta+\text{C}$
$=\sin^{-1}\text{x}+\text{C}$ $(\because\text{ x}=\sin\theta)$
View full question & answer→Question 562 Marks
Evaluate the following integrals:
$\int\limits^2_02\text{x}\big[\text{x}\big]\text{dx}$
Answer[x] = 0 for 0
and [x] = 1 for 1
Hence,
$=\int\limits^1_10+\int\limits^2_12\text{x}\text{ dx}$
$=\big\{\text{x}^2\big\}^2_1$
$=3$
View full question & answer→Question 572 Marks
Write a value of $\int\sin^3\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^3\text{ dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\sin^4\text{x}}{4}+\text{C}$ $(\because\text{t}=\sin\text{x})$
View full question & answer→Question 582 Marks
Evaluate the integral in Exercise:
$\int_{0}^{1}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
$\text{put}\ \text{x}^{2}+1=\text{y},\ \ \therefore 2\text{x}\ \text{dx}=\text{dy},\ \text{or}\ \text{x}\ \text{dx}=\frac{1}{2}\text{dy}$
$\text{when}\ \text{x}=2,\ \text{y}=5$
$\text{when}\ \text{x}=3,\ \text{y}=10$
$\therefore |=\frac{1}{2}\int^{10}_{5}\frac{\text{dy}}{\text{y}}=\frac{1}{2}\big[\log\text{y}\big]^{10}_{5}=\frac{1}{2}\big[\log10-\log5\big]=\frac{1}{2}\log\bigg[\frac{10}{5}\bigg]=\frac{1}{2}\log2$
View full question & answer→Question 592 Marks
Find the integrals of the functions in Exercises:
$\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
Answer$\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
$=\frac{\sin^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}+\frac{\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\cos^2\text{x}}+\frac{\cos\text{x}}{\sin^2\text{x}}$
$=\tan\text{x}\sec\text{x}+\cot\text{x}\text{ cosec x}$
$\therefore\ \ \ \int\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\int\big(\tan\text{x}\sec\text{x}+\cot\text{x}\text{ cosec x}\big)\text{dx}$
$=\sec\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 602 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
Putting $\sqrt{\text{x}}+1=\text{t}$
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{\text{x}}}\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{1}{\text{t}}\text{dt}$
$=2\text{ In }|\text{t}|+\text{C}$
$=2\text{ In }|\sqrt{\text{x}}+1|+\text{C }\big[\because\text{t}=\sqrt{\text{x}}+1\big]$
View full question & answer→Question 612 Marks
Evalute the following integrals:
$\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
Putting $\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{1+\text{t}}$
$=-\text{ln}|1+\text{t}|+\text{C}$
$=-\text{ln}|1+\cot\text{x}|+\text{C}\ \big[\because\text{t}=\cot\text{x}\big]$
View full question & answer→Question 622 Marks
Evaluate the following integrals:
$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
Answer$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{dx}\\-6\int\text{cosec}^2\text{x}+\int\tan^2\text{x dx}-\int\cot^2\text{x dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{x dx}\\-6\int\text{cosec}^2\text{x}+\int(\sec^2\text{x}-1)\text{dx}-\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+6\int\sec^2\text{x dx}-7\int\text{cosec}^2\text{x dx}$
$=-3\cos\text{x}-4\sin\text{x}+6\tan\text{x}+7\cot\text{x}+\text{C}$
View full question & answer→Question 632 Marks
Integrate the function in Exercise:
$\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$
AnswerLet $\text{I}=\int\text{e}^\text{x}(\sin\text{x}+\cos\text{x})\text{dx}$
Let $\text{f}(\text{x})=\sin\text{x}$
$\Rightarrow \ \text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}$
It is known that, $=\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\text{e}^\text{x}\sin\text{x}+\text{C}$
View full question & answer→Question 642 Marks
$\int\tan^2(2\text{x}-3)\text{dx}$
Answer$\int\tan^2(2\text{x}-3)\text{dx}$
$=\int[\sec^2(2\text{x}-3)-1]\text{dx}$
$=\int\sec^2(2\text{x}-3)\text{dx}-\int1\text{dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
View full question & answer→Question 652 Marks
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{\sqrt{16-\text{e}^{2\text{x}}}}\text{ dx}$
Answer$\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$ Let $\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}\text{dx}}{\sqrt{16-(\text{e}^{\text{x}})^2}}$$=\int\frac{\text{dt}}{\sqrt{16-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{4^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{e}^\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 662 Marks
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{\sqrt{4+\tan^2\text{x}}}\text{ dx}$
AnswerLet $\tan\text{x}=\text{t}$ $\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$$=\log\Big|\text{t}+\sqrt{\text{t}^2+4}\Big|+\text{C}$
$=\log\Big|\tan+\sqrt{\tan^2\text{x}+4}\Big|+\text{C}$
View full question & answer→Question 672 Marks
$\int\sin^2\frac{\text{x}}{2}\text{dx}$
AnswerLet I $=\int\sin^2\frac{\text{x}}{2}\text{dx}.$ Then,
$\text{I}=\frac{1}{2}\int2\sin^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int(1-\cos\text{x})\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\frac{1}{2}\int\text{dx}-\frac{1}{2}\int\cos\text{xdx}$
$=\frac{1}{2}\times\text{x}-\frac{1}{2}\times\sin\text{x}+\text{C}$
$=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
$\therefore\text{I}=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
View full question & answer→Question 682 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$ $\int\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}=\sin^{-1}\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\text{I}=\text{F}(1)-\text{F}(0)$ $=\sin^{-1}(1)-\sin^{-1}(0)$ $=\frac{\pi}{2}-0$$=\frac{\pi}{2}$
View full question & answer→Question 692 Marks
Evaluate the following integrals:
$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
Answer$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
$=\int\limits^2_0\sqrt{2^2-\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}+\frac{1}{2}\times2^2\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}\Big]^2_0+2\Big[\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=0+2\Big(\frac{\pi}{2}-0\Big)$
$=\pi$
View full question & answer→Question 702 Marks
Evaluate the definite integral in Exercise:
$\int_{2}^{3}\frac{1}{\text{x}}\text{dx}$
Answer$\text{Let} \ \text{I}=\int\limits_{2}^{3}\frac{1}{\text{x}}\ \text{dx}$$\int\frac{1}{\text{x}}\text{dx}=\text{log}|\text{x|}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(3)-\text{F}(2)$
$=\text{log|3|}-\text{log|2|}=\text{log}\frac{3}{2}$
View full question & answer→Question 712 Marks
Write a value of $\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{t}^3}{1}\text{dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$\text{I}=\frac{(\tan^{-1}\text{x})^4}{4}+\text{C}$
View full question & answer→Question 722 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{1}_{0}\text{x}(1-\text{x})^{\text{n}}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\text{x}(1-\text{x})^{\text{n}}\ \text{dx}=\int\limits_{0}^{1}(1-\text{x)}\left\{1-(1-\text{x)}^{\text{n}}\right\}\text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \text{I}=\int\limits_{0}^{1}(1-\text{x})(1-1+\text{x})^{\text{n}}\ \text{dx}=\int\limits_{0}^{1}(1-\text{x})\text{x}^{\text{n}}\ \text{dx}=\int^{1}_{0}(\text{x}^{\text{n}}-\text{x}^{\text{n}+1})\text{dx}$
$\Rightarrow\ \ \text{I}=\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}-\frac{\text{x}^{\text{n}+2}}{\text{n}+2}\bigg)^{1}_{0}=\frac{1}{\text{n}+1}-\frac{1}{\text{n}+2}-(0-0)=\frac{\text{n}+2-\text{n}-1}{(\text{n}+1)(\text{n}+2)}=\frac{1}{\text{(n}+1)(\text{n}+2)}$
View full question & answer→Question 732 Marks
Evaluate the definite integral in Exercise:$\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\text{n}}{2}}\cos2\text{x}\ \text{dx}$$\int\cos2\text{x}\ \text{dx}=\bigg(\frac{\sin2\text{x}}{2}\bigg)=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{2}\bigg)-\text{F}(0)$
$=\frac{1}{2}\bigg[\sin2\bigg(\frac{\pi}{2}\bigg)-\sin0\bigg]$
$=\frac{1}{2}[\sin\pi-\sin0]$
$=\frac{1}{2}[0-0]=0$
View full question & answer→Question 742 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}=2\int\limits^\text{a}_0\text{f}\big(\text{x}^2\big)\text{dx}$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}$
Here, $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=\text{f}(-\text{x})^2=\text{f}(\text{x})^2=\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x}^2)\text{dx}$ $\bigg[\text{Using}\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx}\text{ when g(x) is even}\bigg]$
View full question & answer→Question 752 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\cos^2\text{x}}{2\sin^2\text{x}}}\text{dx}$
$=\int\cot\text{x dx}$
$=\text{ln}|\sin\text{x}|+\text{C}$
View full question & answer→Question 762 Marks
Write a value of $\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sec^2\text{x}}{(5+\tan\text{x})^4}\text{ dx}$
Let $5+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}^4}$
$=-\frac{1}{3\text{t}^3}+\text{C}$
$\text{I}=-\frac{1}{3(5+\tan\text{x})^3}+\text{C}$
View full question & answer→Question 772 Marks
Evaluate the following integrals:$\int\text{x}^{\text{n}}.\log\text{x dx}$
Answer$\int\text{x}^{\text{n}}\log\text{x dx}$
Taking log x as the first function and $x^n$ as the second function.
$=\log\text{x}\int\text{x}^\text{n}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}\log\text{x}\int\text{x}^\text{n}\text{dx}\Big)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{1}{\text{x}}\bigg(\frac{\text{x}^{n+1}}{\text{n}+1}\bigg)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}}}{\text{n}+1}\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}+1}}{(\text{n}+1)^2}+\text{C}$
View full question & answer→Question 782 Marks
Write the coefficient a, b, c of which the value of the integral $\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$ is independent.
Answer$\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$
$=\Big[\text{a}\frac{\text{x}^3}{3}+\text{b}\frac{\text{x}^2}{2}+\text{cx}\Big]^3_{-3}$
$=9\text{a}+\frac{9}{2}\text{b}+3\text{c}+9\text{a}-\frac{9}{2}\text{b}+3\text{c}$
$=18\text{a}+6\text{c}$
Hence, the given integral is independent of b.
View full question & answer→Question 792 Marks
Evaluate the following integrals:
$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
Answer$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\int\limits^{4}_2\frac{2\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\times\Big[\log(\text{x})^2-1\Big]^4_2$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\frac{1}{2}\big(\log17-\log5\big)$
$=\frac{1}{2}\log\Big(\frac{17}{5}\Big)$ $\Big(\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big)$
View full question & answer→Question 802 Marks
Evaluate the following integrals:
$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$
Answer$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$
$=\big[\tan^{-1}\text{x}\big]^1_0$
$=\tan^{-1}1-\tan^{-1}0$
$=\frac{\pi}{4}-0$
$=\frac{\pi}{4}$
View full question & answer→Question 812 Marks
Evaluate the following:
$\int\frac{\text{dt}}{\sqrt{3\text{t}-2\text{t}^2}}$
AnswerLet $\text{I}=\int\frac{\text{dt}}{\sqrt{3\text{t}-2\text{t}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{-\Big(\text{t}^2-\frac{3}{2}\text{t}\Big)}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\Big(\frac{3}{4}\Big)^2-\Big(\text{t}-\frac{3}{4}\Big)^2}}$
$=\frac{1}{\sqrt{2}}\sin^{-1}\bigg(\frac{\text{t}-\frac{3}{4}}{\frac{3}{4}}\bigg)+\text{C}$ $=\frac{1}{\sqrt{2}}\sin^{-1}\Big(\frac{4\text{t}-3}{3}\Big)+\text{C}$
View full question & answer→Question 822 Marks
Evaluate the following integrals:
$\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
$=\int\limits^{1}_02^{\text{x}-0}\text{ dx}$ $\big(\because[\text{x}]=0,\text{ where}, 0<\text{x}<1\big)$
$=\int\limits^{1}_02^{\text{x}}\text{ dx}$
$=\Big[\frac{2^{\text{x}}}{\log_\text{e}2}\Big]^1_0$
$=\frac{2^1}{\log_\text{e}2}-\frac{2^0}{\log_\text{e}2}$
$=\frac{2}{\log_\text{e}2}-\frac{1}{\log_\text{e}2}$
$=\frac{1}{\log_\text{e}2}$
View full question & answer→Question 832 Marks
Evaluate the following integrals:
$\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
Let $\cos^2\text{x}=\text{t}$
On differentiating both sides, we get
$-2\cos\text{x}\sin\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I}=\int\text{e}^\text{t}2\sin\text{x}\cos\text{x}\frac{\text{dt}}{-2\sin\text{x}\cos\text{x}}$
$=-\int\text{e}^\text{t}\text{dt}$
$=-\text{e}^\text{t}+\text{C}$
$=-\text{e}^{\cos^2\text{x}}+\text{C}$
View full question & answer→Question 842 Marks
If $\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8,$ find the value of a.
AnswerWe have,
$\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8$
$\Rightarrow\Big[3-\frac{\text{x}^3}{3}\Big]^{\text{a}}_0=8$
$\Rightarrow\text{a}^3=8$
$\Rightarrow\text{a}=2$
View full question & answer→Question 852 Marks
Evaluate:
$\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-2\sin^2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{c}$
View full question & answer→Question 862 Marks
Evaluate the following integrals:
$\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$Now, $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
$=2\int\cos\text{t dt}$
$=2\sin\text{t}+\text{C}$
$=2\sin\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 872 Marks
Evaluate the definite integral in Exercise:$\int\limits_0^{\frac{\pi}{4}}\sin2\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\text{x}}{4}}\sin2\text{x} \ \text{dx}$$\int\sin2\text{x}\ \text{dx}=\bigg(\frac{-\cos2\text{x}}{2}\bigg)=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$
$=-\frac{1}{2}\pi\bigg[\cos2\bigg(\frac{\pi}{4}\bigg)-\cos0\bigg] $
$=-\frac{1}{2}\bigg[\cos\bigg(\frac{\pi}{2}\bigg)-\cos0\bigg]$
$=-\frac{1}{2}[0-1]$
$=\frac{1}{2}$
View full question & answer→Question 882 Marks
Evaluate:
$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
$\int\sqrt{\frac{2\sin^2\text{x}}{2}}\text{dx}\ \ [\therefore1-\cos2\text{x}=2\sin^2\text{x]}$
$=\int\sin\text{x dx}$
$=-\cos\text{x}+\text{c}$
View full question & answer→Question 892 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Let $\text{e}^{\sqrt{\text{x}}}=\text{t}$
$\Rightarrow\text{e}^{\sqrt{\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{e}^{\sqrt{\text{x}}}}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
Now, $\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
$=2\int\cos\text{t}\text{ dt}$
$=2\sin\text{t}+\text{C}$
$=2\sin\Big(\text{e}^\sqrt{\text{x}}\Big)+\text{C}$
View full question & answer→Question 902 Marks
Evaluate $\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
$=\int\frac{2}{\sin^2\text{x}+\cos^2\text{x}-(\cos^2\text{x}-\sin^2\text{x})}\text{ dx}$
$=\int\frac{2}{2\sin^2\text{x}}\text{ dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{ dx}$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{C}$
View full question & answer→Question 912 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos2\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos(2\text{x})}\text{dx}$ $\Big[\therefore\ 1-\cos\text{A}=2\sin^2\Big(\frac{\text{A}}{2}\Big)\Big]$
$=\int\frac{\text{dx}}{2\sin^2\text{x}}$
$=\frac{1}{2}\int\text{cosec}^2\text{x dx}$
$=\frac{1}{2}[-\cot\text{x}]+\text{C}$
$=-\frac{1}{2}\cot\text{x}+\text{C}$
View full question & answer→Question 922 Marks
Evalute the following integrals:
$\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
Putting $\text{x}+\cos\text{x}=\text{t}$
$\Rightarrow1-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln t}+\text{C}$
$=\text{ln}|\text{x}+\cos\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\cos\text{x}\big]$
View full question & answer→Question 932 Marks
Evaluate the following integrals:
$\int\cos^{-1}(\sin\text{x})\text{dx}$
Answer$\int\cos^{-1}(\sin\text{x})\text{dx}$
$=\int\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big)\text{dx}$ $\Big[\therefore\ \sin\text{x}=\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$=\int\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 942 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{4}_0\sin2\text{x dx}$
Answer$\int\limits^\frac{\pi}{4}_0\sin2\text{x dx}$
$=\Big[\frac{-\cos2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=-\frac{1}{2}\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$=-\frac{1}{2}\times(0-1)$
$=\frac{1}{2}$
View full question & answer→Question 952 Marks
Evaluate $\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
Let $\text{I}=\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
$=\int\text{e}^{\text{t}}\text{dt}$
$=\text{e}^{\text{t}}+\text{C}$
$=\text{e}^{\tan^{-1}}\text{x}+\text{C}$
View full question & answer→Question 962 Marks
Evaluate $\int\frac{1}{\text{x}^2+16}\text{ dx}$
AnswerSince, $\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}$
Thus, $\int\frac{1}{\text{x}^2+16}\text{ dx}=\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 972 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big(\frac{\text{a}^2}{\text{b}^2}-\text{x}^2}\Big)}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\big(\frac{\text{a}}{\text{b}}\big)^2-\text{x}^2}}$
$=\frac{1}{\text{b}}\sin^{-1}\Big(\frac{\text{xb}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 982 Marks
Evaluate the definite integral in Exercise:
$\int^{\pi}_{0}(\sin^{2}\frac{\text{x}}{2}-\cos^{2}\frac{\text{x}}{2})\text{dx}$
Answer$\text{Let}\text{I}=\int\limits_{0}^{\pi}\bigg(\sin^{2}\frac{\text{x}}{2}-\cos^{2}\frac{\text{x}}{2}\bigg)\text{dx}$$=-\int\limits_{0}^{\pi}\bigg(\cos^{2}\frac{\text{x}}{2}-\sin^{2}\frac{\text{x}}{2}\bigg)\text{dx}$
$=-\int\limits_{0}^{\pi}\cos\text{x}\ \text{dx}$
$\int\cos\text{x}\text{dx}=\sin\text{x}=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(\pi)-\text{F}(0)$
$=\sin\pi-\sin0$
$=0$
View full question & answer→Question 992 Marks
Evaluate the following integrals:
$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Answer$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
$=\int\limits^{\text{e}^2}_\text{e}\frac{\frac{1}{\text{x}}}{\log\text{x}}\text{ dx}$
$=\log\big[(\log{\text{x}})\big]^{\text{e}^2}_\text{e}$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\log\big(\log\text{e}^2\big)-\log(\log\text{e})$
$=\log(2\log\text{e})-\log(\log\text{e})$
$=\log2-\log1$ $(\log\text{e}=1)$
$=\log2-0$
$=\log2$
View full question & answer→Question 1002 Marks
$\int\sin^2\text{bx dx}$
Answer$\int\sin^2\text{bx dx}$
$=\int\Big[\frac{1-\cos2\text{bx}}{2}\Big]\text{dx}$ $\Big[\therefore\sin^2\text{x}=\frac{1-\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos2\text{bx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{bx}}{2\text{b}}\Big]+\text{C}$
$=\frac{\text{x}}{2}-\frac{\sin2\text{bx}}{4\text{b}}+\text{C}$
View full question & answer→Question 1012 Marks
Verify the following:
$\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}=\log\big|\text{x}^2+3\text{x}\big|+\text{c}$
AnswerLet $\text{I}=\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}$
Put $\text{x}^2+3\text{x}=\text{t}$
$\Rightarrow\ (2\text{x}+3)\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}}\text{dt}=\log|\text{t}|+\text{C}$ $=\log\big|(\text{x}^2+3\text{x})\big|+\text{C}$
View full question & answer→Question 1022 Marks
Integrate the function in Exercise:
$\text{x} \ \sec^2\text{x}$
AnswerLet $\text{I}=\int\text{x}\sec^2\text{x dx}$
Taking x as first function and $sec^2x$ as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sec^2\text{x dx}-\int\Big[\Big\{\frac{\text{d}}{\text{dx}}\text{x}\int\sec^2\text{x} \ \text{dx}\Big\}\text{dx}\Big]$
$=\text{x}\tan\text{x}-\int1.\tan\text{x dx}$
$=\text{x}\tan\text{x}+\text{log}|\cos\text{x}|+\text{C}$
View full question & answer→Question 1032 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
$= \int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi}{2}+0+\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 1042 Marks
Evaluate the following integrals:
$\int\limits^2_0\big[\text{x}\big]\text{dx}$
Answerwe have,
$\text{I}=\int\limits^2_0\big[\text{x}\big]\text{dx}$
$=\int\limits^1_0\big[\text{x}\big]\text{dx}+\int\limits^2_1\big[\text{x}\big]\text{dx}$
$=\int\limits^1_00\text{ dx}+\int\limits^2_1(\text{1})\text{dx}$ $\begin{bmatrix}\because\big[\text{x}\big]=\begin{cases}0,&0\leq\text{x}<1\\1,&1\leq\text{x}<2\end{cases}\end{bmatrix}$
$=0+\big[\text{x}\big]^{2}_1$
$=2-1$
$=1$
View full question & answer→Question 1052 Marks
Evalute the following integrals:
$\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
Putting $\text{x}+\log\text{x}=\text{t}$
$\Rightarrow1+\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{x}+1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\text{x}|+\text{C}$
View full question & answer→Question 1062 Marks
Evaluate the following integral:
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
Answer$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dx}}{\text{a}^2-\text{b}^2\text{x}}$
$=\frac{1}{\text{b}^2}\times\frac{1}{2\frac{\text{a}}{\text{b}}}\log\Bigg|\frac{\frac{\text{a}}{\text{b}}+\text{x}}{\frac{\text{a}}{\text{b}}-\text{x}}\Bigg|+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big|+\text{C}\Big]$
$=\frac{1}{2\text{ab}}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}\Big|+\text{C}$
View full question & answer→Question 1072 Marks
Write a value of $\int\text{a}^{\text{x}}\text{e}^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{a}^{\text{x}}\text{e}^{\text{x}}\text{ dx}$$=\int(\text{a}\text{e})^{\text{x}}\text{ dx}$
$=\frac{(\text{a}\text{e})^{\text{x}}}{\log\text{ae}}+\text{C}$
$\therefore\ \text{I}=\frac{(\text{a}\text{e})^{\text{x}}}{\log\text{ae}}+\text{C}$
View full question & answer→Question 1082 Marks
Integrate the functions in Exercises:
$\frac{3\text{x}^2}{\text{x}^6+1}$
Answer$\text{Let I}=\int\frac{3\text{x}^2}{\text{x}^6+1}\text{ dx}$
$=\int\frac{3\text{x}^2}{(\text{x}^3)^2+1}\text{ dx} \ \ \ \ \ ...\text{(i)}$
Putting$\ \ \ \text{x}^3=\text{t} \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2=\frac{\text{dt}}{\text{dx}} \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt}$
$\therefore \ \ \ \ \ $From eq. (i),$\ \ \ \ \text{I}=\int\frac{\text{dt}}{\text{t}^2+1}=\frac{1}{1}\tan^{-1}\frac{\text{t}}{1}+\text{c}$
$=\tan^{-1}\text{x}^3+\text{c}$
View full question & answer→Question 1092 Marks
If $\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$ and $\text{f}'(1)=\frac{1}{2}$, find f'(x).
Answer$\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$
$\text{f}'\text{(x)}=\text{x}-\text{x}^{-2}$
$\int\text{f}'\text{(x)}\text{dx}=\int(\text{x}-\text{x}^{-2})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}-\frac{\text{x}^{-2+1}}{-2+1}+\text{C}$
$=\frac{\text{x}^2}{2}+\frac{1}{\text{x}}+\text{C}$
$\text{f}'\text{(1)}=\frac{1}{2}$ (Given)
$\Rightarrow\frac{{1}^2}{2}+\frac{1}{1}+\text{C}=\frac{1}{2}$
$\Rightarrow\text{C}=-1$
$\therefore\ \text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{1}{\text{x}}-1$
View full question & answer→Question 1102 Marks
Evaluate the following integrals:
$\int\Big(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\Big)^2\text{dx}$
Answer$\int\Big(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\Big)^2\text{dx}$
$=\int\Big(\text{x}+\frac{1}{\text{x}}-2\Big)\text{dx}$
$=\int\text{xdx}+\int\frac{\text{dx}}{\text{x}}-2\int\text{dx}$
$=\frac{\text{x}^2}{2}+\ln|\text{x}|-2\text{x}+\text{C}$
View full question & answer→Question 1112 Marks
Evaluate the following integrals:
$\int\text{x}\text{ cosec}^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x cosec}^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\text{cosec}^2\text{x dx}-\int(\int\text{cosec}^2\text{x dx})\text{dx}$
$=-\text{x}\cot\text{x}+\int\cot\text{x dx}$
$=-\text{x}\cot\text{x}+\log|\sin\text{x}|+\text{C}$
View full question & answer→Question 1122 Marks
Evaluate the following integrals:
$\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
We know that,
$\text{x}[\text{x}]=\begin{cases}\text{x}\times0,&0<\text{x}<1\\\text{x}\times1,&1<\text{x}<2\end{cases}$
i.e.,
$\text{x}[\text{x}]=\begin{cases}0,&0<\text{x}<1\\\text{x},&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0\text{x}[\text{x}]\text{dx}+\int\limits^{2}_1\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0{0}\text{ dx}+\int\limits^{2}_1(\text{x})\text{dx}$
$=0+\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=\frac{2^2}{2}-\frac{1^2}{2}$
$=\frac{3}{2}$
View full question & answer→Question 1132 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{1+\tan\text{x}}{\text{x}+\log\sec\text{x}}\text{dx}$
Putting $\text{x}+\log\sec\text{x}=\text{t}$
$\Rightarrow1+\frac{\sec\text{x}\tan\text{x}}{\sec\text{c}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\tan\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sec\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\log\sec\text{x}\big]$
View full question & answer→Question 1142 Marks
Evaluate the following integrals:
$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Answer$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{1}{3}\Big[\tan^{-1}\frac{\text{x}}{3}\Big]^3_0$
$=\frac{1}{3}\big(\tan^{-1}1-\tan^{-1}0\big)$
$=\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{\pi}{12}$
View full question & answer→Question 1152 Marks
Integrate the following integrals:
$\int\sin\text{mx}\cos\text{nx dx m}\neq\text{n}$
Answer$\int\sin(\text{mx})\cos(\text{nx) dx}$
$=\frac{1}{2}\int2\sin(\text{mx})\cos(\text{nx})\text{dx}$
$=\frac{1}{2}\int[\sin(\text{mx}+\text{nx})+\sin(\text{mx}-\text{nx})]\text{dx}$ $[\therefore2\sin\text{A}\cos\text{B}=\sin(\text{A}+\text{B})+\sin(\text{A}-\text{B})]$
$=\frac{1}{2}\Big[-\frac{\cos(\text{m+n})\text{x}}{\text{m}+\text{n}}-\frac{\cos(\text{m}-\text{n})\text{x}}{\text{m}-\text{n}}\Big]+\text{C}$
View full question & answer→Question 1162 Marks
Evaluate the following integrals:
$\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
AnswerLet $\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
We have,
$|\text{x}|=\begin{cases}\text{x},&0\leq\text{x}\leq1\\-\text{x},&-2\leq\text{x}<0\end{cases}$
$\therefore\ \frac{|\text{x}|}{\text{x}}=\begin{cases}1,&0\leq\text{x}\leq1\\-1,&-2\leq\text{x}<0\end{cases}$
Therefore,
$\text{I}=\int\limits^0_{-2}-1\text{ dx}+\int\limits^1_01\text{ dx}$
$=-\big[\text{x}\big]^0_{-2}+\big[\text{x}\big]_0^1$
$=0-2+1-0$
$=-1$
View full question & answer→Question 1172 Marks
Evaluate the following integrals:
$\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}=\frac{1}{2}\int\limits^1_0\text{e}^{\text{x}^2}2\text{dx}$
Putting $\text{x}^2=\text{z}$
$2\text{x dx}=\text{dz}$
When $\text{x}\rightarrow0;\text{ z}\rightarrow0$
And $\text{x}\rightarrow1;\text{ z}\rightarrow1$
$\therefore\ \text{I}=\frac{1}{2}\int\limits^1_0\text{e}^2\text{ dz}$
$=\frac{1}{2}\times\big[\text{e}^{\text{z}}\big]^1_0$
$=\frac{1}{2}(\text{e}-\text{e}^0)$
$=\frac{1}{2}(\text{e}-1)$
View full question & answer→Question 1182 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{4}}\tan\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^\frac{\pi}{4}\tan\text{x}\ \text{dx}$ $\int\tan\text{x}\ \text{dx}=-\text{log}|\cos\text{x|}=\text{F}\text{(x)}$By second fundamental theoram of calculus, we obtain
$\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$ $=-\text{log}|\cos\frac{\pi}{4}|+\text{log}|\cos0|$ $=-\text{log}\big|\frac{1}{\sqrt{2}}\big|+\text{log}|1|$ $=\text{log}(2)^\frac{1}{2}$$=\frac{1}{2}\text{log}2$
View full question & answer→Question 1192 Marks
Prove the following Exercise:
$\int^{1}_{-1}\text{x}^{17}\cos^{4}\text{x}\ \text{dx}=0$
Answer$\text{Let I}=\int^{1}\limits_{-1}\text{x}^{17}\cos^{4}\text{x dx}$
Also, let $\text{f(x)}=\text{x}^{17}\cos^{4}\text{x}$
$\Rightarrow\text{f}\ (-\text{x)}=(-\text{x)}^{17}\cos^{4}(-\text{x)}=-\text{x}^{17}\cos^{4}\text{x}=-\text{f (x)}$
Therefore, f(x) is an odd function.
it is know that if f(x) is an odd function, then $\int^{\text{a}}\limits_{-\text{a}}\text{f (x) dx}=0$
$\therefore\text{I}=\int^{1}\limits_{-1}\text{x}^{17}\cos^{4}\text{x dx}=0$
Hence, the given result is Proved.
View full question & answer→Question 1202 Marks
Evaluate $\int\frac{\text{x}^2+4\text{x}}{\text{x}^3+6\text{x}^2+5}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+4\text{x}}{\text{x}^3+6\text{x}^2+5}\text{ dx}$
Let $\text{x}^3+6\text{x}^2+5=\text{t}$
$(3\text{x}^2+12\text{x})\text{dx}=\text{dt}$
$3(\text{x}^2+4\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{3}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{3}\log\big|\text{x}^3+6\text{x}^2+5\big|+\text{C}$
View full question & answer→Question 1212 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}\tan\text{x}}{3\sec\text{x}+5}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}\tan\text{x}}{3\sec\text{x}+5}\text{dx}$ then,
Putting $\sec\text{x}=\text{t}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\sec\text{x}\tan\text{x}$
$\Rightarrow\text{dt}=\sec\text{x}\tan\text{x dx}$
$\therefore\text{I}=\int\frac{\text{dt}}{3\text{t}+5}$
$=\frac{1}{3}\text{ln}|3\text{t}+\text{5}|+\text{C}$
$=\frac{1}{3}\text{ln}|3\sec\text{x}+5|+\text{C}\ \big[\because\text{t}=\sec\text{x}\big]$
View full question & answer→Question 1222 Marks
Evaluate the following integrals:
$\int\bigg\{\text{x}^2+\text{e}^{\log\text{x}}+\Big(\frac{\text{e}}{2}\Big)^\text{x}\bigg\}\text{dx}$
Answer$\int\bigg\{\text{x}^2+\text{e}^{\log\text{x}}+\Big(\frac{\text{e}}{2}\Big)^\text{x}\bigg\}\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{e}^{\log\text{x}}\text{dx}+\int\Big(\frac{\text{e}}{2}\Big)^\text{x}\text{dx}$
$=\frac{\text{x}^3}{3}+\int\text{xdx}+\int\Big(\frac{\text{e}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}+\frac{1}{\log\big(\frac{\text{e}}{2}\big)}\times\Big(\frac{\text{e}}{2}\Big)^\text{x}+\text{C}$
View full question & answer→Question 1232 Marks
Evaluate the following integrals:
$\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$ Now, $\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$$=2\int\sin\text{t}\text{ dt}$
$=2[-\cos\text{t}]+\text{C}$
$=-2\cos\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 1242 Marks
Evaluate the definite integral in Exercise:$\int\limits_{0}^{1}\frac{\text{dx}}{1+\text{x}^{2}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{4}\frac{\text{dx}}{1+\text{x}^{2}}$
$\int\frac{\text{dx}}{1+\text{x}^{2}}=\tan^{-1}\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain$\text{I}=\text{F}(1)-\text{F}(0)$
$=\tan^{-1}(1)-\tan^{-1}(0)$
$=\frac{\pi}{4}$
View full question & answer→Question 1252 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\Big(\frac{\text{x}-1}{2\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{x}\frac{1}{2\text{x}}\text{dx}-\int\text{e}^{\text{x}}\frac{1}{2\text{x}^2}\text{dx}$
Integration by parts
$=\frac{\text{e}^{\text{x}}}{2\text{x}}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2\text{x}}\Big)\Big)\text{dx}-\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{2\text{x}}+\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}-\int\frac{\text{e}^{\text{x}}}{2\text{x}^2}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{2\text{x}}+\text{C}$
View full question & answer→Question 1262 Marks
Evaluate the following integrals:
$\int\text{x}^{\frac{5}{4}}\text{dx}$
Answer$\int\text{x}^{\frac{5}{4}}\text{dx}=\frac{\text{x}^{\frac{5}{4}}+1}{\frac{5}{4}+1}+\text{c}$
$=\frac{\text{x}^{\frac{5+4}{4}}+\text{c}}{\frac{5+4}{4}}$
$=\frac{4\text{x}^{\frac{9}{4}}}{9}+\text{c}$
View full question & answer→Question 1272 Marks
Evaluate the following integrals:
$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
Answer$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\cot^{-1}\Big(\frac{2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$ $\big[\therefore\ \sin2\text{x}=2\sin\text{x}\cos\text{x} \text{ & }1-\cos2\text{x}=2\sin^2\text{x}\big]$
$=\int\cot^{-1}(\cot\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1282 Marks
Evaluate $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{\text{dx}}{2\sqrt{\text{x}}}=\text{dt}$
$\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$
Putting $\sqrt{\text{x}}=\text{t}$ and $\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$
$\therefore\ \text{I}=2\int\sec^2+\text{dt}$
$=2\tan\text{t}+\text{C}$
$=2\tan(\sqrt{\text{x}})+\text{C}$ $(\because\text{t}=\sqrt{\text{x}})$
View full question & answer→Question 1292 Marks
Evaluate the following integrals:
$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
$\text{Let},1+\sqrt{\text{x}}=\text{t}$
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$
$\text{Now},\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
$=2\int\text{t}^2\text{dt}$
$=\frac{2}{3}\text{t}^3+\text{C}$
$=\frac{2}{3}(1+\sqrt{\text{x}})^3+\text{C}$
View full question & answer→Question 1302 Marks
Integrate the function in Exercise:
$\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}$
Answer$\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}=\frac{\text{e}^{4\log\text{x}}\big(\text{e}^{\log\text{x}}-1\big)}{\text{e}^{2\log\text{x}}\big(\text{e}^{\log\text{x}}-1\big)}$$=\text{e}^{2\log\text{x}}$
$=\text{e}^{\log\text{x}^{2}}$
$=\text{x}^{2}$
$\therefore\int\frac{\text{e}^{5\log\text{x}}-\text{e}^{4\log\text{x}}}{\text{e}^{3\log\text{x}}-\text{e}^{2\log\text{x}}}\text{dx}=\int\text{x}^{2}\text{dx}=\frac{\text{x}^{3}}{3}+\text{C}$
View full question & answer→Question 1312 Marks
Write a value of $\int\text{e}^{\text{ax}}\sin\text{bx}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{ax}}\sin\text{bx}\text{ dx}$
We know that,
$\int\text{e}^{\text{ax}}\sin\text{bx}\text{ dx}=\frac{\text{e}^{\text{ax}}}{\text{a}^2+\text{b}^2}\big[\text{a}\sin\text{bx}-\text{b}\cos\text{bx}\big]+\text{C}$
Thus,
$\text{I}=\frac{\text{e}^{\text{ax}}}{\text{a}^2+\text{b}^2}\big[\text{a}\sin\text{bx}-\text{b}\cos\text{bx}\big]+\text{C}$
View full question & answer→Question 1322 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos\text{x}\text{ dx}$
AnswerWe have,
$\int\text{x}^2\cos\text{x dx}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x})\big(\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int\sin\text{x }2\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[-\text{x}\cos\text{x}+\int\cos\text{x dx}\big]$
$\therefore\ \int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos\text{x}\text{ dx}=\big[\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=\Big[\frac{\pi}{4}+0-2-0-0+0\Big]$
$=\frac{\pi^2}{4}-2$
View full question & answer→Question 1332 Marks
If $\int\limits^{\text{a}}_03\text{x}^2\text{ dx}=8,$ Write the value of a.
AnswerWe have,
$\int\limits^{\text{a}}_03\text{x}^2\text{ dx}=8$
$\Rightarrow\Big[3\frac{\text{x}^3}{3}\Big]^{\text{a}}_0=8$
$\Rightarrow\big[\text{x}^3\big]^{\text{a}}_0=8$
$\Rightarrow\text{a}^3-0=8$
$\Rightarrow\text{a}=\sqrt[3]{8}$
$\Rightarrow\text{a}=2$
View full question & answer→Question 1342 Marks
Evaluate $\int\frac{\text{x}+\cos6\text{x}}{3\text{x}^2+\sin6\text{x}}\text{ dx}$
AnswerLet $3\text{x}^2+\sin6\text{x}=\text{t}$
$6\text{x}+6\cos6\text{x dx}=\text{dt}$
$(\text{x}+\cos6\text{x})\text{dx}=\frac{\text{dt}}{6}$
Thus, $\text{I}=\int\frac{\text{x}+\cos6\text{x}}{3\text{x}^2+\sin6\text{x}}\text{ dx}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{6}\log|\text{t}|+\text{C}$
$=\frac{\log\big|3\text{x}^2+\sin6\text{x}\big|}{6}+\text{C}$
View full question & answer→Question 1352 Marks
Evaluate the following integrals:
$\int\log\text{x}\frac{\sin\big\{1+(\log\text{x})^2\big\}}{\text{x}}\text{ dx}$
Answer$\int\log\text{x}\frac{\sin\big\{1+(\log\text{x})^2\big\}}{\text{x}}\text{ dx}$
Let $1+(\log\text{x})^2=\text{t}$
$\Rightarrow2\log\text{x}\times\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\log\text{x}}{\text{x}}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\log\text{x}\frac{\sin\big\{1+(\log\text{x})^2\big\}}{\text{x}}\text{ dx}$
$=\frac{1}{2}\int\sin(\text{t})\text{dt}$
$=\frac{1}{2}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{2}\cos\big\{1+(\log\text{x})^2\big\}+\text{C}$
View full question & answer→Question 1362 Marks
Evalute the following integrals:
$\int\frac{1-\sin2\text{x}}{\text{x}+\cos^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\sin2\text{x}}{\text{x}+\cos^2\text{x}}\text{dx}$
Putting $\text{x}+\cos^2\text{x}=\text{t}$
$\Rightarrow1-2\cos\text{x}\times\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1-\sin2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\text{x}+\cos^2\text{x}|\text{C}\ \big[\because\text{t}=\text{x}+\cos^2\text{x}\big]$
View full question & answer→Question 1372 Marks
Integrate the rational function in exercise:
$\frac{1}{\text{x}^2-9}$
Answer$\frac{1}{\text{x}^2-9}\text{dx}$
$=\int\frac{1}{\text{x}^2-3^2}\text{dx}$
$=\frac{1}{2\times3}\text{log}\big|\frac{\text{x}-3}{\text{x}+3}\big|+\text{c}$
$\bigg[\therefore \ \int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\text{log}\big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\big|\bigg]$
$=\frac{1}{6}\text{log}\big|\frac{\text{x}-3}{\text{x}+3}\big|+\text{c}$
View full question & answer→Question 1382 Marks
Evaluate the following integrals:
$\int\Big(\frac{\text{m}}{\text{x}}+\frac{\text{x}}{\text{m}}+\text{m}^\text{x}+\text{x}^\text{m}+\text{mx}\Big)\text{dx}$
Answer$\int\Big(\frac{\text{m}}{\text{x}}+\frac{\text{x}}{\text{m}}+\text{m}^\text{x}+\text{x}^\text{m}+\text{mx}\Big)\text{dx}$
$=\text{m}\int\frac{1}{\text{x}}\text{dx}+\frac{1}{\text{m}}\int\text{xdx}+\int\text{m}^\text{x}\text{dx}\int\text{x}^\text{m}\text{dx}+\text{m}\int\text{xdx}$
$=\text{m}\log|\text{x}|+\frac{\text{x}^2}{2\text{m}}+\frac{\text{m}^\text{x}}{\log\text{m}}+\frac{\text{x}^{\text{m}+1}}{\text{m}+1}+\frac{\text{mx}^2}{2}+\text{C}$
View full question & answer→Question 1392 Marks
Write a value of $\int\text{e}^{\text{ax}}\{\text{a }\text{f(x)}+\text{f}'(\text{x})\}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{ax}}\{\text{a }\text{f(x)}+\text{f}'(\text{x})\}\text{ dx}$
We know that,
$\int\text{e}^{\text{ax}}\{\text{a }\text{f(x)}+\text{f}'(\text{x})\}\text{ dx}=\text{e}^{\text{ax}}\text{f(x)}+\text{C}$
Which is a general formula.
View full question & answer→Question 1402 Marks
Integrate the function in Exercise:
$\frac{\text{x}^{3}}{\sqrt{1-\text{x}^{8}}}$
Answer$\frac{\text{x}^{3}}{\sqrt{1-\text{x}^{8}}}$ $\text{Let}\ \text{x}^{4}=\text{t}\Rightarrow4\text{x}^{3}\text{dx}=\text{dt}$ $\Rightarrow\int\frac{\text{x}^{3}}{\sqrt{1-\text{x}^{8}}}\text{dx}=\frac{1}{4}\int\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$ $=\frac{1}{4}\sin^{-1}\text{t}+\text{C}$$=\frac{1}{4}\sin^{-1}\text{(x}^{4})+\text{C}$
View full question & answer→Question 1412 Marks
Write a value of $\int\sqrt{\text{x}^2-9}\text{ dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2-9}\text{ dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{ dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
Here $a = 3, a^2 = 9$
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{\text{x}^2-9}-\frac{9}{2}\log\Big|\text{x}+\sqrt{\text{x}^2+9}\Big|+\text{C}$
View full question & answer→Question 1422 Marks
Integrate the following integrals:
$\int\cos\text{mx}\cos\text{nx dx m}\neq\text{n}$
Answer$\int\cos\text{mx}\cos\text{nx dx}$
$=\frac{1}{2}\int2\cos(\text{mx})\cos(\text{nx})\text{dx}$
$=\frac{1}{2}\int[\cos(\text{mx}+\text{nx})+\cos(\text{mx}-\text{nx})]\text{dx}$ $[\therefore2\cos\text{A}\cos\text{B}=\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})]$
$=\frac{1}{2}\Big[\frac{\sin(\text{m+n})\text{x}}{\text{m}+\text{n}}+\frac{\sin(\text{m}-\text{n})\text{x}}{\text{m}-\text{n}}\Big]+\text{C}$
View full question & answer→Question 1432 Marks
Evalute the following integrals:
$\int\frac{1}{\sin\text{x}\cos^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin\text{x}\cos^2\text{x}}\text{dx},$ then,
$\text{I}=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^2}{\sin\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\sec\text{x}\tan\text{x }\text{dx}+\int\text{cosec x dx}$
$=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
$\therefore\text{I}=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1442 Marks
Integrate the function in Exercise:$\frac{2+\sin2\text{x}}{1+\cos2\text{x}}\text{e}^{\text{x}}$
Answer$\text{I}=\int\bigg(\frac{2+\sin2\text{x}}{1+\cos2\text{x}}\bigg)\text{e}^{\text{x}}$
$=\int\bigg(\frac{2+2\sin\text{x}\cos\text{x}}{2\cos^{2}\text{x}}\bigg)\text{e}^{\text{x}}$
$=\int\bigg(\frac{1+\sin\text{x}\cos\text{x}}{\cos^{2}\text{x}}\bigg)\text{e}^{\text{x}}$
$=\int\big(\sec^{2}\text{x}+\tan\text{x}\big)\text{e}^{\text{x}}$
$\text{Let}\ \text{f(x)}=\tan\text{x}\Rightarrow\text{f(x)}=\sec^{2}\text{x}$
$\therefore\text{I}=\int\big[\text{f(x)}+\text{f}'\text{(x)}\big]\text{e}^{\text{x}}\text{dx}$
$=\text{e}^{\text{x}}\ \text{f(x)}+\text{C}$
$=\text{e}^{\text{x}}\tan\text{x}+\text{C}$
View full question & answer→Question 1452 Marks
Evaluate the following integrals:
$\int\limits^\text{x}_{0}\text{e}^{-\text{x}}\text{ dx}$
Answer$\int\limits^\text{x}_{0}\text{e}^{-\text{x}}\text{ dx}$
$=-\big[\text{e}^{-\text{x}}\big]^{\text{x}}_0$
$=-(0-1)$
$=0+1$
$=1$
View full question & answer→Question 1462 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
AnswerLet $\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Here, $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
Consider, $\text{f}(-\theta)=\log\bigg[\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\bigg]$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)=-\text{f}(\theta)$
i.e., $\text{f}(\theta)$ is odd function.
Therefore, $\text{I}=0$
View full question & answer→Question 1472 Marks
Evaluate the following definite integrals:
$\int_{-2}\limits^{\frac{1}{2}}\frac{1}{\sqrt{1-\text{x}^2}} \text{ dx}$
AnswerLet $\int_{-2}\limits^{\frac{1}{2}}\frac{1}{\sqrt{1-\text{x}^2}} \text{ dx}$ Then,
$\text{I}=\big[\sin^{-1}\text{x}\big]^{\frac{1}{2}}_0$
$\Rightarrow\text{I}=\sin^{-1}\frac{1}{2}-\sin^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{6}-0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
View full question & answer→Question 1482 Marks
Evaluate $\int\cos^{-1}(\sin\text{x})\text{dx}$
Answer$\int\cos^{-1}(\sin\text{x})\text{dx}=\int\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big)\text{dx}$
$=\int\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\frac{\pi}{2}\text{x}-\frac{1}{2}\text{x}^2+\text{C}$
Hence, $\int\cos^{-1}(\sin\text{x})\text{dx}=\frac{\pi}{2}\text{x}-\frac{1}{2}\text{x}^2+\text{C}$
View full question & answer→Question 1492 Marks
Evaluate the following:
$\int\frac{(\text{x}^2+2)}{\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+2}{\text{x}+1}\text{dx}$$=\int\Big(\text{x}-1+\frac{3}{\text{x}+1}\Big)\text{dx}$
$=\int(\text{x}-1)\text{dx}+3\int\frac{1}{\text{x}+1}\text{dx}$
$=\frac{\text{x}^2}{2}-\text{x}+3\log\big|(\text{x}+1)\big|+\text{C}$
View full question & answer→Question 1502 Marks
Evaluate the following integrals:
$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$
Answer$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$
$=\int\text{x}^\text{e}\text{dx}+\int\text{e}^\text{x}\text{dx}+\text{e}^\text{e}\int1\text{dx}$
$=\frac{\text{x}^{\text{e}+1}}{\text{e}+1}+\text{e}^\text{x}+\text{x}.\text{e}^\text{e}+\text{C}$
View full question & answer→Question 1512 Marks
Evaluate $\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
AnswerLet $\sqrt{\text{x}}=\text{t}$
$\text{x}=\text{t}^2$
$1-\text{x}=1-\text{t}^2$
$-\text{dx}=-2\text{tdt}$
$\text{dx}=2\text{tdt}$
$\text{I}=\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int(1-\text{t}^2)\text{t}2\text{t dt}$
$=2\int (1-\text{t}^2)\text{t}^2\text{ dt}$
$=2\big(\int\text{t}^2\text{dt}-\int\text{t}^4\text{dt}\big)$
$=2\frac{\text{t}^3}{3}-2\frac{\text{t}^5}{5}+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}+\text{C}$
View full question & answer→Question 1522 Marks
Evaluate the integral in Exercise:$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{1+\cos^{2}\text{x}}\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{1+\cos^{2}\text{x}}\text{dx}$
$\text{put}\ \cos\text{x}=\text{t},\therefore\sin\text{x}\ \text{dx}=\text{dt}\Rightarrow\sin\text{x}\ \text{dx}=-\text{dt}\ \text{when}\ \text{x}=0,\text{t}=\cos0=1$
$\text{when}\ \text{x}=\frac{\pi}{2},\text{t}=\cos\frac{\pi}{2}=0$
$\therefore\ \ \text{I}=-\int\limits^{0}_{1}\frac{\text{dt}}{1+\text{t}^{2}}=-\big[\tan^{-1}\text{t}\big]^{0}_{1}=-\big[\tan^{-1}0-\tan^{-1}1\big]=-\bigg[0-\frac{\pi}{4}\bigg]=\frac{\pi}{4}$
View full question & answer→Question 1532 Marks
$\int\cos^2\text{nx dx}$
Answer$\int\cos^2\text{nx dx}$
$=\int\Big[\frac{1+\cos2\text{nx}}{2}\Big]\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1+\cos2\text{nx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}+\frac{\sin2\text{nx}}{2\text{n}}\Big]+\text{C}$
$=\frac{\text{x}}{2}+\frac{\sin2\text{nx}}{4\text{n}}+\text{C}$
View full question & answer→Question 1542 Marks
Integrate the functions in Exercises:
$\frac{1}{\sqrt{(2-\text{x})^2}+1}$
Answer$\int\frac{1}{\sqrt{(2-\text{x})^2}+1}\text{ dx}$
$=\frac{\log\bigg|(2-\text{x})+\sqrt{(2-\text{x})^2+1^2}\bigg|}{-1\rightarrow\text{Coeff. of x}}+\text{c}$$ \ \ \ \ \ \ \ \bigg[\because\int\frac{1}{\sqrt{\text{x}^2+\text{a} ^2}}\text{ dx}=\log\bigg|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\bigg|\bigg]$
$=-\log\bigg|2-\text{x}+\sqrt{4+\text{x}^2-4\text{x}+1}\bigg|+\text{c}$
$=\log\Bigg|\frac{1}{2-\text{x}\sqrt{\text{x}^2-4\text{x}+5}}\Bigg|+\text{C}$
View full question & answer→Question 1552 Marks
Evaluate the following integrals:
$\int\limits^3_2\frac{1}{\text{x}}\text{ dx}$
Answer$\int\limits^3_2\frac{1}{\text{x}}\text{ dx}$
$=\Big[\log_\text{e}\text{x}\Big]^3_2$
$=\log_\text{e}3-\log_\text{e}2$
$=\log_\text{e}\Big(\frac{3}{2}\Big)$
View full question & answer→Question 1562 Marks
Evaluate the following integrals:$\int\text{xe}^{2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\text{xe}^{2\text{x}}\text{dx}$
Using integration by parts,
$\text{I}=\text{x}\int\text{e}^{2\text{x}}\text{dx}-\int(1\times\int\text{e}^{2\text{x}}\text{dx})\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\int\Big(\frac{\text{e}^{2\text{x}}}{2}\Big)\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\frac{\text{e}^{2\text{x}}}{4}+\text{C}$
$\text{I}=\Big(\frac{\text{x}}{2}-\frac{1}{4}\Big)\text{e}^{2\text{x}}+\text{C}$
View full question & answer→Question 1572 Marks
Integrate the function in Exercise:$\text{f}\big(\text{ax}+\text{b}\big)\big[\text{f(ax}+\text{b)}\big]^{\text{n}}$
Answer$\text{f' (ax}+\text{b)}\big[\text{f (ax}+\text{b)}\big]^{\text{n}}$ $\text{Let}\ \text{f (ax}+\text{b)}=\text{t}\Rightarrow \text{af}\ \text{(ax}+\text{b)}\text{dx}=\text{dt}$ $\Rightarrow\int\text{f}\ \big(\text{ax}+\text{b}\big)\big[\text{f}\text{(ax}+\text{b)}\big]^{\text{n}}\text{dx}=\frac{1}{\text{a}}\int\text{t}^{\text{n}}\text{dt}$$=\frac{1}{\text{a}}\bigg[\frac{\text{t}^{\text{n}+1}}{\text{n}+1}\bigg]$
$=\frac{1}{\text{a}(\text{n}+1)}\big(\text{f (ax}+\text{b)}\big)^{\text{n+1}}+\text{C}$
View full question & answer→Question 1582 Marks
Evaluate the following integrals:
$\int\frac{1}{3\sqrt{\text{x}^2}}\text{dx}$
Answer$\int\frac{\text{dx}}{3\sqrt{\text{x}^2}}$
$=\int\frac{\text{dx}}{\text{x}^\frac{2}{3}}$
$=\int\text{x}^\frac{-2}{3}\text{dx}$
$=\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\text{c}$
$=3\text{x}^\frac{1}{3}+\text{c}$
View full question & answer→Question 1592 Marks
Evaluate the following integrals:
$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$
Answer$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$ Let $\sin^{-1}\text{x}=\text{t}$ $\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ Now, $\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$$=\int\text{t}^3\text{ dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{(\sin^{-1}\text{x})^4}{4}+\text{C}$
View full question & answer→Question 1602 Marks
Evaluate the following integrals:
$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$
Answer$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$
$=\int\limits^4_0\frac{1}{\sqrt{14^2-\text{x}^2}}\text{ dx}$
$=\Big[\sin^{-1}\frac{\text{x}}{4}\Big]^4_0$
$=\Big(\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 1612 Marks
Evaluate the following definite integrals:
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
AnswerWe know that $\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{C}$
Now,
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
$=\big[\log(\text{x}+7)\big]^3_{-2}$
$=\big[\log10-\log5]^3_{-2}$
$=\log\frac{10}{5}$ $\Big[\because\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big]$
$=\log2$
$\therefore\ \int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}=\log2$
View full question & answer→Question 1622 Marks
Evaluate the following integrals:
$\int3^{2\log_3\text{x}}\text{dx}$
Answer$\int3^{2\log_3\text{x}}\text{dx}=\int3^{\log_3\text{x}^2}\text{dx}$
$=\int\text{x}^2\text{dx}$
$=\frac{\text{x}^3}{3}+\text{c}$
$\int\log_\text{x}\text{xdx}=\int1\text{dx}$
$=\text{x}+\text{c}$
View full question & answer→Question 1632 Marks
Write the primitive or anti-derivative of $\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$.
Answer$\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$
Integrating both sides:
$\int\text{f(x)}\text{dx}=\int\Big(\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}\Big)\text{dx}$
$=\bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]+\bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}^{\frac{1}{2}}+\text{C}$
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