Question 512 Marks
Evaluate the following integrals:
$\int\text{e}^{-2\text{x}}\sin\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{-2\text{x}}\sin\text{x }\text{dx}$
$\because\ \int\text{e}^{2\text{x}}\sin\text{bx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\sin\text{bx}-\text{b}\cos\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^{-2\text{x}}}{5}\{-2\sin\text{x}-\cos\text{x}\}+\text{C}$
View full question & answer→Question 522 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 532 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 542 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}}(3-5\text{x})\text{dx}$
Answer$\int\sqrt{\text{x}}(3-5\text{x})\text{dx}$
$=\int\text{x}^{\frac{1}{2}}(3-5\text{x})\text{dx}$
$=\int\Big(3\text{x}^\frac{1}{2}-5\text{x}^\frac{3}{2}\Big)\text{dx}$
$=3\bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]-5\bigg[\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\bigg]+\text{C}$
$=2\text{x}^{\frac{3}{2}}-2\text{x}^{\frac{5}{2}}+\text{C}$
View full question & answer→Question 552 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 562 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 572 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 582 Marks
Evaluate $\int\frac{\log\text{x}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$=\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \text{I}=\int\text{t dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{1}{2}(\log\text{x})^2+\text{C}$
View full question & answer→Question 592 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 602 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 612 Marks
Evaluate the following intregals:
$\int\frac{6\text{x}-5}{\sqrt{3\text{x}^2-5\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{6\text{x}-5}{\sqrt{3\text{x}^2-5\text{x}+1}}\text{ dx}$
putting $3\text{x}^2-5\text{x}+1=\text{t}$
$\Rightarrow(6\text{x}-5)\ \text{dx}=\text{dt}$
Then,
$\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{3\text{x}^2-5\text{x}+1}+\text{C}$
View full question & answer→Question 622 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^5+\text{x}^{-2}+2}{\text{x}^2}\text{dx}$
Answer$\int\frac{\text{x}^5+\text{x}^{-2}+2}{\text{x}^2}\text{dx}$
$=\int\bigg(\frac{\text{x}^5}{\text{x}^2}+\frac{\text{x}^{-2}}{\text{x}^2}+\frac{2}{\text{x}^2}\bigg)\text{dx}$
$=\int\text{x}^3\text{dx}+\int \text{x}^{-4}+2\int\text{x}^{-2}\text{dx}$
$=\frac{\text{x}^4}{4}+\frac{\text{x}^{-3}}{-3}+\frac{2\text{x}^{-1}}{-1}+\text{C}$
$=\frac{\text{x}^4}{4}-\frac{\text{x}^{-3}}{-3}-\frac{2}{\text{x}}+\text{C}$
View full question & answer→Question 632 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 642 Marks
Evaluate the following intregals:
$\int\frac{\text{x}-1}{\sqrt{\text{x}^2+1}}\text{dx}$
Answer$\int\frac{\text{x}-1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\int\frac{\text{x}}{\sqrt{\text{x}^2+1}}-\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2+1}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\frac{1}{2}(2\sqrt{\text{t}})-\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}\\=\sqrt{\text{t}}-\text{In}\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{c}$
$=\sqrt{\text{x}^2+1}-\text{In}\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{c}$
View full question & answer→Question 652 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 662 Marks
Evaluate:
$\int\frac{1}{\text{a}^\text{x}\text{b}^\text{x}}\text{dx}$
Answer$\int\frac{1}{\text{a}^\text{x}\text{b}^\text{x}}\text{dx}=\int\text{a}^{-\text{x}}\text{b}^{-\text{x}}\text{dx}$
$=\int(\text{ab})^{-\text{x}}\text{dx}$
$=\frac{(\text{ab})^{-\text{x}}}{\log_\text{e}(\text{ab})^{-1}}+\text{c}$
$=\frac{(\text{ab})^{-\text{x}}}{-\log_\text{e}(\text{ab})}+\text{c}$
$=\frac{\text{a}^{-\text{x}}\text{b}^{-\text{x}}}{-\log_\text{e}(\text{ab})}+\text{c}$
View full question & answer→Question 672 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 682 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 692 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 702 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 712 Marks
Evaluate the following integrals:
$\int\sqrt{3-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{3-\text{x}^2}\text{dx}$
$=\int\sqrt{(\sqrt3)^2-\text{x}^2}\text{dx}$
$\text{I}=\frac{\text{x}}{2}\sqrt{3-\text{x}^2}+\frac{3}{2}\sin^{-1}\Big(\frac{\text{x}}{\sqrt3}\Big)+\text{C}$
View full question & answer→Question 722 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 732 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 742 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 752 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 762 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 772 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 782 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 792 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
Answer$\int\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$ Let $\tan^{-1}\text{x}=\text{t}$ $\Rightarrow\Big(\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{dt}$ Now, $\int\Big(\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1+\text{x}^2}\Big)\text{ dx}$ $=\int\text{e}^{\text{mt}}\text{dt}$
$=\frac{\text{e}^{\text{mt}}}{\text{m}}+\text{C}$
$=\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{\text{m}}+\text{C}$
View full question & answer→Question 802 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 812 Marks
Evaluate the following integrals:
$\int\frac{\log\text{x}}{\text{x}}\text{dx}$
Answer$\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$\text{Let},\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\text{Now},\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$=\int\text{t}\text{dt}$
$=\frac{\text{t}^2}{2}+\text{C}=\frac{(\log\text{x})^2}{2}+\text{C}$
View full question & answer→Question 822 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 832 Marks
Evaluate the following integrals:
$\int\frac{1}{1+\text{x}-\text{x}^2}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\text{x}-\text{x}^2}$
$=\int\frac{-\text{dx}}{\text{x}^2-\text{x}-1}$
$=\int\frac{-\text{dx}}{\text{x}^2-\text{x}+\frac{1}{4}-\frac{1}{4}-1}$
$=\int\frac{-\text{dx}}{\big(\text{x}-\frac{1}{2}\big)^2-\frac{5}{4}}$
$=\int\frac{\text{dx}}{\frac{5}{4}-\big(\text{x}-\frac{1}{2}\big)^2}$
View full question & answer→Question 842 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 852 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, a2 = 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 862 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 872 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 882 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 892 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 902 Marks
Evaluate $\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx}$
Let $1+\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{(1+\log\text{x})^3}{3}+\text{C}$
View full question & answer→Question 912 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 922 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 932 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 942 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 952 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 962 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 972 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 982 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big({\frac{\text{a}^2}{\text{b}^2}+\text{x}^2\Big)}}}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\text{x}^2+\big(\frac{\text{a}}{\text{b}}\big)^2}}$
$=\frac{1}{\text{b}}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{\text{a}^2}{\text{b}^2}}\Big|+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\text{x}+\frac{\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\frac{\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\log\text{b}\Big]+\text{C}$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\frac{\log\text{b}}{\text{b}}+\text{C}$
Let $\text{C}-\frac{\log\text{b}}{\text{b}}+\text{C}'$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|+\text{C}'$
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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 1002 Marks
Evalute the following integrals:
$\int\frac{\text{e}^\text{x}+1}{\text{e}^\text{x}+\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^\text{x}+1}{\text{e}^\text{x}+\text{x}}\text{dx}\ .....(\text{i})$
let ex + x = t then,
d(ex + x) = dt
⇒ (ex + x)dx = dt
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{e}^\text{x}+1}$
Putting ex + x = t and $\text{dx}=\frac{\text{dt}}{\text{e}^\text{x}+1}$ in equation (i), we get,
$\text{I}=\int\frac{\text{e}^\text{x}+1}{\text{e}}\times\frac{\text{dt}}{\text{e}^\text{x}+1}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{e}^\text{x}+\text{x}|+\text{C}$
$\therefore\text{I}=\log|\text{e}^\text{x}+\text{x}|+\text{C}$
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