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
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}}$$=\int\frac{\text{dt}}{\sqrt{16-\text{t}^2}}$
$=\int\frac{\text{dt}}{\sqrt{4^2-\text{t}^2}}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{e}^\text{x}}{4}\Big)+\text{C}$
View full question & answer→Question 102 Marks
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{\sqrt{4+\tan^2\text{x}}}\text{ dx}$
AnswerLet $\tan\text{x}=\text{t}$ $\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+2^2}}$$=\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 112 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 122 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 132 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 142 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 152 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 162 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}$
View full question & answer→Question 172 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 182 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 192 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 202 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 212 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 222 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 232 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 242 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 252 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 262 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 272 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 282 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 292 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 302 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 312 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 322 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 332 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 342 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 352 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 362 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 372 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 382 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 392 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 402 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 412 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 422 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 432 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 442 Marks
Evaluate the following integrals:
$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
Answer$\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$\text{Let }1+\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\text{Now,}\int\sqrt{1+\text{e}^\text{x}}\text{ e}^\text{x}\text{dx}$
$=\int\sqrt{\text{t}}\text{ dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{2}{3}(1+\text{e}^\text{x})^\frac{3}{2}+\text{C}$
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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, $\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^\text{x}}{\text{x}}+\text{C}$
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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 472 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 482 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 $\sin\text{x}$
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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 502 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,
$\text{I}=\big[-\cos\text{x}+\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+1-(-1+0)$
$\Rightarrow\text{I}=2$
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