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$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$
$=\int\text{x}^\text{e}\text{dx}+\int\text{e}^\text{x}\text{dx}+\text{e}^\text{e}\int1\text{dx}$
$=\frac{\text{x}^{\text{e}+1}}{\text{e}+1}+\text{e}^\text{x}+\text{x}.\text{e}^\text{e}+\text{C}$
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