Class 12: Indefinite Integrals

$\displaystyle \textbf{Question 1: }~\int\left((2x-3)^5+\sqrt{3x+2}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int (2x-3)^5\,dx+\int (3x+2)^{1/2}\,dx$
$\displaystyle = \int (2x-3)^5\,dx+\frac{1}{3}\int (3x+2)^{1/2}\,d(3x+2)$
$\displaystyle = \frac{1}{2}\int (2x-3)^5\,d(2x-3)+\frac{1}{3}\cdot\frac{(3x+2)^{1/2+1}}{1/2+1}+C$
$\displaystyle = \frac{1}{2}\cdot\frac{(2x-3)^{5+1}}{5+1}+\frac{1}{3}\cdot\frac{(3x+2)^{3/2}}{3/2}+C$
$\displaystyle = \frac{(2x-3)^6}{12}+\frac{2}{9}(3x+2)^{3/2}+C$
$\\$

$\displaystyle \textbf{Question 2: }~\int\left(\frac{1}{(7x-5)^3}+\frac{1}{\sqrt{5x-4}}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int (7x-5)^{-3}\,dx+\int (5x-4)^{-1/2}\,dx$
$\displaystyle = \frac{1}{7}\int (7x-5)^{-3}\,d(7x-5)+\frac{1}{5}\int (5x-4)^{-1/2}\,d(5x-4)$
$\displaystyle = \frac{1}{7}\cdot\frac{(7x-5)^{-3+1}}{-3+1}+\frac{1}{5}\cdot\frac{(5x-4)^{-1/2+1}}{-1/2+1}+C$
$\displaystyle = \frac{1}{7}\cdot\frac{(7x-5)^{-2}}{-2}+\frac{1}{5}\cdot\frac{(5x-4)^{1/2}}{1/2}+C$
$\displaystyle = -\frac{1}{14(7x-5)^2}+\frac{2}{5}\sqrt{5x-4}+C$
$\\$

$\displaystyle \textbf{Question 3: }~\int\left(\frac{1}{2-3x}+\frac{1}{\sqrt{3x-2}}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int (2-3x)^{-1}\,dx+\int (3x-2)^{-1/2}\,dx$
$\displaystyle = \int \frac{1}{2-3x}\,dx+\frac{1}{3}\int (3x-2)^{-1/2}\,d(3x-2)$
$\displaystyle = -\frac{1}{3}\int \frac{1}{2-3x}\,d(2-3x)+\frac{1}{3}\cdot\frac{(3x-2)^{-1/2+1}}{-1/2+1}+C$
$\displaystyle = -\frac{1}{3}\log|2-3x|+\frac{1}{3}\cdot\frac{(3x-2)^{1/2}}{1/2}+C$
$\displaystyle = -\frac{1}{3}\log|2-3x|+\frac{2}{3}\sqrt{3x-2}+C$
$\\$

$\displaystyle \textbf{Question 4: }~\int\frac{x+3}{(x+1)^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{(x+1)+2}{(x+1)^4}\,dx$
$\displaystyle = \int \left((x+1)^{-3}+2(x+1)^{-4}\right)\,dx$
$\displaystyle = \int (x+1)^{-3}\,dx+2\int (x+1)^{-4}\,dx$
$\displaystyle = \frac{(x+1)^{-3+1}}{-3+1}+2\cdot\frac{(x+1)^{-4+1}}{-4+1}+C$
$\displaystyle = \frac{(x+1)^{-2}}{-2}+2\cdot\frac{(x+1)^{-3}}{-3}+C$
$\displaystyle = -\frac{1}{2(x+1)^2}-\frac{2}{3(x+1)^3}+C$
$\\$

$\displaystyle \textbf{Question 5: }~\int\frac{1}{\sqrt{x+1}+\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}\,dx$
$\displaystyle = \int \frac{\sqrt{x+1}-\sqrt{x}}{(x+1)-x}\,dx$
$\displaystyle = \int \left(\sqrt{x+1}-\sqrt{x}\right)\,dx$
$\displaystyle = \int (x+1)^{1/2}\,dx-\int x^{1/2}\,dx$
$\displaystyle = \frac{(x+1)^{1/2+1}}{1/2+1}-\frac{x^{1/2+1}}{1/2+1}+C$
$\displaystyle = \frac{2}{3}(x+1)^{3/2}-\frac{2}{3}x^{3/2}+C$
$\\$

$\displaystyle \textbf{Question 6: }~\int\frac{1}{\sqrt{2x+3}+\sqrt{2x-3}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{\sqrt{2x+3}-\sqrt{2x-3}}{(\sqrt{2x+3}+\sqrt{2x-3})(\sqrt{2x+3}-\sqrt{2x-3})}\,dx$
$\displaystyle = \int \frac{\sqrt{2x+3}-\sqrt{2x-3}}{(2x+3)-(2x-3)}\,dx$
$\displaystyle = \int \frac{\sqrt{2x+3}-\sqrt{2x-3}}{6}\,dx$
$\displaystyle = \frac{1}{6}\int (2x+3)^{1/2}\,dx-\frac{1}{6}\int (2x-3)^{1/2}\,dx$
$\displaystyle = \frac{1}{6}\cdot\frac{1}{2}\int (2x+3)^{1/2}\,d(2x+3)-\frac{1}{6}\cdot\frac{1}{2}\int (2x-3)^{1/2}\,d(2x-3)$
$\displaystyle = \frac{1}{12}\cdot\frac{(2x+3)^{1/2+1}}{1/2+1}-\frac{1}{12}\cdot\frac{(2x-3)^{1/2+1}}{1/2+1}+C$
$\displaystyle = \frac{1}{12}\cdot\frac{(2x+3)^{3/2}}{3/2}-\frac{1}{12}\cdot\frac{(2x-3)^{3/2}}{3/2}+C$
$\displaystyle = \frac{1}{18}(2x+3)^{3/2}-\frac{1}{18}(2x-3)^{3/2}+C$
$\\$

$\displaystyle \textbf{Question 7: }~\int\frac{2x}{(2x+1)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{(2x+1)-1}{(2x+1)^2}\,dx$
$\displaystyle = \int \left(\frac{1}{2x+1}-\frac{1}{(2x+1)^2}\right)\,dx$
$\displaystyle = \int \frac{1}{2x+1}\,dx-\int (2x+1)^{-2}\,dx$
$\displaystyle = \frac{1}{2}\int \frac{1}{2x+1}\,d(2x+1)-\frac{1}{2}\int (2x+1)^{-2}\,d(2x+1)$
$\displaystyle = \frac{1}{2}\log|2x+1|-\frac{1}{2}\cdot\frac{(2x+1)^{-2+1}}{-2+1}+C$
$\displaystyle = \frac{1}{2}\log|2x+1|+\frac{1}{2(2x+1)}+C$
$\\$

$\displaystyle \textbf{Question 8: }~\int\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a}-\sqrt{x+b})}\,dx$
$\displaystyle = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}\,dx$
$\displaystyle = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}\,dx$
$\displaystyle = \frac{1}{a-b}\int (x+a)^{1/2}\,dx-\frac{1}{a-b}\int (x+b)^{1/2}\,dx$
$\displaystyle = \frac{1}{a-b}\cdot\frac{(x+a)^{1/2+1}}{1/2+1}-\frac{1}{a-b}\cdot\frac{(x+b)^{1/2+1}}{1/2+1}+C$
$\displaystyle = \frac{2}{3(a-b)}\left((x+a)^{3/2}-(x+b)^{3/2}\right)+C$
$\\$

$\displaystyle \textbf{Question 9: }~\int \sin x\,\sqrt{1+\cos 2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \sin x\,\sqrt{2\cos^2 x}\,dx \quad (\because~1+\cos 2x=2\cos^2 x)$
$\displaystyle = \int \sin x\cdot \sqrt{2}\cos x\,dx$
$\displaystyle = \sqrt{2}\int \sin x\cos x\,dx$
$\displaystyle = \sqrt{2}\int \frac{1}{2}\sin 2x\,dx$
$\displaystyle = \frac{\sqrt{2}}{2}\int \sin 2x\,dx$
$\displaystyle = \frac{\sqrt{2}}{2}\cdot\left(-\frac{\cos 2x}{2}\right)+C$
$\displaystyle = -\frac{\sqrt{2}}{4}\cos 2x+C$
$\\$

$\displaystyle \textbf{Question 10: }~\int\frac{1+\cos x}{1-\cos x}\,dx. \hspace{5.0cm} \;[\text{CBSE 2000}]$
$\displaystyle \text{Answer:}$
$\displaystyle \int\frac{1+\cos x}{1-\cos x}\,dx = \int\frac{2\cos^2\frac{x}{2}}{2\sin^2\frac{x}{2}}\,dx \quad[\because\;1+\cos x=2\cos^2\tfrac{x}{2},\;1-\cos x=2\sin^2\tfrac{x}{2}]$
$\displaystyle = \int \cot^2\frac{x}{2}\,dx$
$\displaystyle = \int\left(\mathrm{cosec}^2\frac{x}{2}-1\right)\,dx$
$\displaystyle = -2\cot\frac{x}{2}-x+C$

$\\$

$\displaystyle \textbf{Question 11: }~\int\frac{1-\cos x}{1+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \tan^2\!\left(\frac{x}{2}\right)\,dx \quad \left(\because~\frac{1-\cos x}{1+\cos x}=\tan^2\!\left(\frac{x}{2}\right)\right)$
$\displaystyle = \int \left(\sec^2\!\left(\frac{x}{2}\right)-1\right)\,dx$
$\displaystyle = 2\tan\!\left(\frac{x}{2}\right)-x+C$
$\\$

$\displaystyle \textbf{Question 12: }~\int\frac{1}{1-\sin\left(\frac{x}{2}\right)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{1+\sin\left(\frac{x}{2}\right)}{1-\sin^2\left(\frac{x}{2}\right)}\,dx$
$\displaystyle = \int \frac{1+\sin\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}\,dx$
$\displaystyle = \int \left(\sec^2\left(\frac{x}{2}\right)+\sec\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)\right)\,dx$
$\displaystyle = 2\tan\left(\frac{x}{2}\right)+2\sec\left(\frac{x}{2}\right)+C$
$\\$

$\displaystyle \textbf{Question 13: }~\int\frac{1}{1+\cos 3x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int\frac{1}{1+\cos 3x}\,dx = \int\frac{1-\cos 3x}{(1+\cos 3x)(1-\cos 3x)}\,dx$
$\displaystyle = \int\frac{1-\cos 3x}{1-\cos^2 3x}\,dx = \int\frac{1-\cos 3x}{\sin^2 3x}\,dx$
$\displaystyle = \int\left(\mathrm{cosec}^2 3x-\mathrm{cosec}\,3x\cot 3x\right)\,dx$
$\displaystyle = \int \mathrm{cosec}^2 3x\,dx-\int \mathrm{cosec}\,3x\cot 3x\,dx$
$\displaystyle = -\frac{\cot 3x}{3}+\frac{\mathrm{cosec}\,3x}{3}+C$
$\displaystyle = \frac{1}{3}\left(\mathrm{cosec}\,3x-\cot 3x\right)+C$
$\displaystyle = \frac{1}{3}\left(\frac{1-\cos 3x}{\sin 3x}\right)+C$
$\\$

$\displaystyle \textbf{Question 14: }~\int (e^x+1)^2\,e^x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int (e^x+1)^2\,d(e^x)$
$\displaystyle = \frac{(e^x+1)^{2+1}}{2+1}+C$
$\displaystyle = \frac{(e^x+1)^3}{3}+C$
$\\$

$\displaystyle \textbf{Question 15: }~\int\left(e^x+\frac{1}{e^x}\right)^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \left(e^x+e^{-x}\right)^2\,dx$
$\displaystyle = \int \left(e^{2x}+2+e^{-2x}\right)\,dx$
$\displaystyle = \int e^{2x}\,dx+2\int 1\,dx+\int e^{-2x}\,dx$
$\displaystyle = \frac{e^{2x}}{2}+2x-\frac{e^{-2x}}{2}+C$
$\\$

$\displaystyle \textbf{Question 16: }~\int\frac{1+\cos 4x}{\cot x-\tan x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{2\cos^2 2x}{\cot x-\tan x}\,dx \quad (\because~1+\cos 4x=2\cos^2 2x)$
$\displaystyle = \int \frac{2\cos^2 2x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}\,dx$
$\displaystyle = \int \frac{2\cos^2 2x}{\frac{\cos^2 x-\sin^2 x}{\sin x\cos x}}\,dx$
$\displaystyle = \int \frac{2\cos^2 2x\sin x\cos x}{\cos 2x}\,dx$
$\displaystyle = \int 2\cos 2x\sin x\cos x\,dx$
$\displaystyle = \int \cos 2x\cdot \sin 2x\,dx$
$\displaystyle = \frac{1}{2}\int \sin 4x\,dx$
$\displaystyle = -\frac{1}{8}\cos 4x+C$
$\\$

$\displaystyle \textbf{Question 17: }~\int\frac{1}{\sqrt{x+3}-\sqrt{x+2}}\,dx. \hspace{5.0cm} \;[\text{CBSE 2002}]$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \frac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3}-\sqrt{x+2})(\sqrt{x+3}+\sqrt{x+2})}\,dx$
$\displaystyle = \int \frac{\sqrt{x+3}+\sqrt{x+2}}{(x+3)-(x+2)}\,dx$
$\displaystyle = \int \left(\sqrt{x+3}+\sqrt{x+2}\right)\,dx$
$\displaystyle = \int (x+3)^{1/2}\,dx+\int (x+2)^{1/2}\,dx$
$\displaystyle = \frac{(x+3)^{1/2+1}}{1/2+1}+\frac{(x+2)^{1/2+1}}{1/2+1}+C$
$\displaystyle = \frac{2}{3}(x+3)^{3/2}+\frac{2}{3}(x+2)^{3/2}+C$
$\\$

$\displaystyle \textbf{Question 18: }~\int \tan^2(2x-3)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle = \int \left(\sec^2(2x-3)-1\right)\,dx$
$\displaystyle = \frac{1}{2}\int \sec^2(2x-3)\,d(2x-3)-\int 1\,dx$
$\displaystyle = \frac{1}{2}\tan(2x-3)-x+C$
$\\$

$\displaystyle \textbf{Question 19: }~\int\frac{1}{\cos^2 x\,(1-\tan x)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int\frac{1}{\cos^2 x(1-\tan x)^2}\,dx$
$\displaystyle = \int\frac{\sec^2 x}{(1-\tan x)^2}\,dx$
$\displaystyle \text{Let } 1-\tan x=t$
$\displaystyle -\sec^2 x\,dx=dt \;\Rightarrow\; \sec^2 x\,dx=-dt$
$\displaystyle \therefore I=\int\frac{-dt}{t^2}$
$\displaystyle = -\int t^{-2}\,dt$
$\displaystyle = -\left(\frac{t^{-1}}{-1}\right)+C$
$\displaystyle = \frac{1}{t}+C$
$\displaystyle = \frac{1}{1-\tan x}+C$