Integration By Parts (Part-4)| S N Dey| Class 12 Examhoop

In the previous article, we have discussed the solutions of VSA type questions of Integration By Parts Chapter of S N Dey mathematics Class 12. In this chapter, we will discuss the problems of Short Answer Type Questions.

Integration By Parts (Part 4), S N Dey Maths

Short Answer Type Questions of Integration By Parts :

$21.~~\displaystyle\int{x^2\cos^2x}~dx$

Solution.

$\displaystyle\int{x^2\cos^2x}~dx\\=\frac 12 \displaystyle\int{x^2(2\cos^2x)}~dx\\=\frac 12 \int{x^2(1+\cos2x)}~dx\\=\frac 12 \int{x^2}~dx+\frac 12 \int{x^2\cos2x}~dx\\=\frac 12 \cdot \frac{x^3}{3}+\frac 12I_1.$

$I_1\\=\displaystyle\int{x^2\cos2x}~dx\\=x^2\int{\cos2x}~dx-\int{\left[\frac{d}{dx}(x^2)\int{\cos2x}~dx\right]}~dx\\=x^2 \cdot (\frac 12 \sin2x)-\int{2x \cdot (\frac 12 \sin2x)}~dx\\=\frac{x^2}{2}\sin2x-\int{x\sin2x~dx}\\=\frac{x^2}{2}\sin2x-I_2$

$I_2\\=\displaystyle\int{x\sin2x}~dx\\=x\int{\sin2x}~dx-\int{\left[\frac{d}{dx}(x)\int{\sin2x}~dx\right]}~dx\\=x \cdot (-\frac 12 \cos2x)-\int{(-\frac 12\cos2x)}~dx\\=-\frac x2 \cos2x+\frac 12 \cdot \frac{\sin2x}{2}$

$\therefore \displaystyle\int{x^2\cos^2x}~dx\\=\frac 12 \cdot \frac{x^3}{3}+\frac 12(\frac{x^2}{2}\sin2x-I_2)\\=\frac{x^3}{6}+\frac{x^2}{4}\sin2x-\frac 12(-\frac x2 \cos2x+\frac 12 \cdot \frac{\sin2x}{2})\\=\frac{x^3}{6}+\frac 18(2x^2\sin2x+2x\cos2x-\sin2x)+c.$

$22.~~\displaystyle\int{\frac{x+\sin x}{1+\cos x}}~dx$

Solution.

$\displaystyle\int{\frac{x+\sin x}{1+\cos x}}~dx\\=\int{\frac{x}{1+\cos x}}~dx+\int{\frac{\sin x}{1+\cos x}}~dx\\=\int{\frac{x}{2\cos^2(x/2)}}~dx+\int{\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}}~dx\\=\frac 12\int{x\sec^2(x/2)}~dx+\int{\tan(x/2)}~dx\\=\frac 12I_1+\int{\tan(x/2)}~dx$

$I_1\\=\displaystyle\int{x\sec^2(x/2)}~dx\\=x\int{\sec^2(x/2)}~dx+\int{\left[\frac{d}{dx}(x)\int{\sec^2(x/2)}~dx\right]}~dx\\=x \cdot 2\tan(x/2)-\int{1 \cdot (2\tan(x/2))}~dx\\=2x\tan(x/2)-2\int{\tan(x/2)}~dx$

$\therefore~ \displaystyle\int{\frac{x+\sin x}{1+\cos x}}~dx\\=\frac 12(2x\tan(x/2)-2\int{\tan(x/2)}~dx)+\int{\tan(x/2)}~dx\\=x\tan(x/2)-\int{\tan(x/2)}~dx+\int{\tan(x/2)}~dx\\=x\tan(x/2)+c.$

$23.~~\displaystyle\int{x^210^{2x}}~dx$

Solution.

$\displaystyle\int{x^210^{2x}}~dx\\=x^2\int{10^{2x}}~dx- \int{\left[\frac{d}{dx}(x^2)\int{10^{2x}}~dx\right]}~dx\\=x^2 \cdot \frac{10^{2x}}{2\log 10}-\int{2x \cdot \frac{10^{2x}}{2\log 10}}~dx\\=x^2 \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{\log 10} \int{x \cdot10^{2x}}~dx\\=x^2 \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{\log 10}I_1.$

$I_1\\=\displaystyle\int{x \cdot10^{2x}}~dx\\=x\int{10^{2x}}~dx-\int{\left[\frac{d}{dx}(x)\int{10^{2x}}~dx\right]}~dx\\=x \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{2\log 10}\int{10^{2x}}~dx\\=x \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{2\log 10} \cdot \frac{10^{2x}}{2\log 10}$

$\therefore~~ \displaystyle\int{x^210^{2x}}~dx\\=x^2 \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{\log 10}\left(x \cdot \frac{10^{2x}}{2\log 10}-\frac{1}{2\log 10} \cdot \frac{10^{2x}}{2\log 10}\right)\\= \frac{10^{2x}}{2\log 10} \left[x^2-\frac{x}{\log 10}+\frac{1}{2(\log 10)^2}\right]+c.$

$24.~~\displaystyle\int{x^2\sin^{-1}(3x)}~dx$

Solution.

$\displaystyle\int{x^2\sin^{-1}(3x)}~dx\\=\sin^{-1}(3x)\int{x^2}~dx-\int{\left[\frac{d}{dx}(\sin^{-1}(3x))\int{x^2}~dx\right]}~dx\\=\frac{x^3}{3}\sin^{-1}(3x)-\int{\frac{3}{\sqrt{1-(3x)^2}}\cdot \frac{x^3}{3}}~dx\\=\frac{x^3}{3}\sin^{-1}(3x)-\int{\frac{x^3}{\sqrt{1-9x^2}}}~dx\\=\frac{x^3}{3}\sin^{-1}(3x)-\int{\frac{x^2 \cdot x}{\sqrt{1-9x^2}}}~dx\\=\frac{x^3}{3}\sin^{-1}(3x)-I_1$

$~\text{let}~~1-9x^2=z^2 \\ \Rightarrow -18x~dx=2z~dz \\ \Rightarrow x~dx=-\frac 19 z~dz$

$\therefore~I_1\\=\displaystyle\int{\frac{x^2 \cdot x}{\sqrt{1-9x^2}}}~dx\\=-\frac 19\int{\frac{1-z^2}{9} \cdot \frac 1z \cdot z}~dz\\=-\frac{1}{81}\left[z-\frac{z^3}{3}\right]\\=-\frac{1}{81}\left[\sqrt{1-9x^2}-\frac{(1-9x^2)\sqrt{1-9x^2}}{3}\right]$

$~\therefore~\displaystyle\int{x^2\sin^{-1}(3x)}~dx\\=\frac{x^3}{3}\sin^{-1}(3x)+\frac{1}{81}\left[\sqrt{1-9x^2}-\frac{(1-9x^2)\sqrt{1-9x^2}}{3}\right]\\=\frac{x^3}{3}\sin^{-1}(3x)+\frac{1}{81}\sqrt{1-9x^2}\left[1-\frac{1-9x^2}{3}\right]\\=\frac{x^3}{3}\sin^{-1}(3x)+\frac{1}{81}\sqrt{1-9x^2} \left[\frac{3-1+9x^2}{3}\right]\\=\frac{x^3}{3}\sin^{-1}(3x)+\frac{1}{243}(2+9x^2)\sqrt{1-9x^2}+c.$

$25.~~\displaystyle\int{(\sin^{-1}x)^2}~dx$

Solution.

$\displaystyle\int{(\sin^{-1}x)^2}~dx\\=(\sin^{-1}x)^2\int{dx}-\int{\left[\frac{d}{dx}(\sin^{-1}x)^2\int{dx}~\right]}~dx\\=x(\sin^{-1}x)^2-\int{2\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}} \cdot x}~dx\\=x(\sin^{-1}x)^2-2I_1.$

$I_1=\displaystyle\int{x\sin^{-1}x \cdot \frac{1}{\sqrt{1-x^2}}}~dx \rightarrow(1)$

$~\text{let}~~\sin^{-1}x=z \Rightarrow \frac{dx}{\sqrt{1-x^2}}=dz \rightarrow(2).$

From $~(1),~(2)~$ we get,

$I_1\\=\displaystyle\int{z\sin z}~dz\\=z\int{\sin z}~dz-\int{\left[\frac{d}{dx}(z)\int{\sin z}~dz\right]}~dz\\=z(-\cos z)+\int{1 \cdot \cos z}~dz\\=-z\cos z+\sin z\=-z\sqrt{1-\sin^2z}+\sin z\=-(\sin^{-1}x))\sqrt{1-x^2}+x$

$\therefore~ \displaystyle\int{(\sin^{-1}x)^2}~dx \\=x(\sin^{-1}x)^2-2\left[-(\sin^{-1}x))\sqrt{1-x^2}+x\right]+c\\=x(\sin^{-1}x)^2+2\sqrt{1-x^2}~\sin^{-1}x-2x+c.$

$26.~~ \displaystyle\int{\sin^{-1}(3x-4x^3)}~dx$

Solution.

$\text{let}~x=\sin\theta . \\ \therefore~\sin^{-1}(3x-4x^3)\\=\sin^{-1}(3\sin\theta-4\sin\theta)\\=\sin^{-1}(\sin3\theta)\\=3\theta\\=3\sin^{-1}x$

$\displaystyle\int{\sin^{-1}(3x-4x^3)}~dx\\=3\int{\sin^{-1}x}~dx=3I_1.$

$I_1\\=\displaystyle\int{\sin^{-1}x}~dx\\=\sin^{-1}x\int{dx}-\int{\left[\frac{d}{dx}(\sin^{-1}x)\int{~dx}\right]}~dx\\=x\sin^{-1}x-\int{\frac{1}{\sqrt{1-x^2}} \cdot x}~dx\\=x\sin^{-1}x-I_2.$

$\text{let}~~z^2=1-x^2 \\ \therefore 2z~dz=-2x~dx \\ \text{or,}~~x~dx=-z~dz$

$I_2\\=\displaystyle\int{\frac{1}{z} \cdot (-z)~dz}\\=-\int{dz}\\=-z\\=-\sqrt{1-x^2}.$

$~\therefore~ \displaystyle\int{\sin^{-1}(3x-4x^3)}~dx\\=3x\sin^{-1}x-3(-\sqrt{1-x^2})+c\\=3x\sin^{-1}x+3\sqrt{1-x^2}+c.$

$27.~~\displaystyle\int{\tan^{-1}\frac{3x-x^2}{1-3x^2}}~dx$

Solution.

$\displaystyle\int{\tan^{-1}\frac{3x-x^2}{1-3x^2}}~dx\\=3\int{\tan^{-1}x}~dx\\=3I_1.$

$I_1\\=\displaystyle\int{\tan^{-1}x}~dx\\=\tan^{-1}x\int{dx}-\int{\left[\frac{d}{dx}(\tan^{-1}x)\int{~dx}\right]}~dx\\=x\tan^{-1}x-\int{\frac{1}{1+x^2} \cdot x~dx}\\=x\tan^{-1}x-\frac 12\int{\frac{2x}{1+x^2}}~dx\\=x\tan^{-1}x-\frac 12\int{\frac{d(1+x^2)}{1+x^2}}\\=x\tan^{-1}x-\frac 12\log|1+x^2|$

$\therefore~\displaystyle\int{\tan^{-1}\frac{3x-x^2}{1-3x^2}}~dx\\=3\left(x\tan^{-1}x-\frac 12\log|1+x^2|\right)+c.$

$28.~~ \displaystyle\int{\frac{\cos^{-1}x}{x^3}}~dx$

Solution.

$I\\=\displaystyle\int{\frac{\cos^{-1}x}{x^3}}~dx\\=\cos^{-1}x\int{\frac{1}{x^2}}~dx-\int{\left[\frac{d}{dx}(\cos^{-1}x)\int{\frac{1}{x^2}}~dx\right]}~dx\\=(\cos^{-1}x) \times \frac{x^{-3+1}}{-3+1}-\int{\frac{-1}{\sqrt{1-x^2}} \times \frac{x^{-3+1}}{-3+1}}~dx\\=-\frac{\cos^{-1}x}{2x^2}-\frac 12\int{\frac{1}{x^2\sqrt{1-x^2}}}~dx\\=-\frac{\cos^{-1}x}{2x^2}-\frac 12I_1 \rightarrow(1)$

$~\text{let}~x=\sin\theta \ \therefore~dx=\cos\theta ~d\theta \rightarrow(2)$

Then, by $~(1),~(2)~$ we get,

$~~~I_1\\=\displaystyle\int{\frac{1}{x^2\sqrt{1-x^2}}}~dx\\=\int{\frac{1}{\sin^2\theta\sqrt{(1-\sin^2\theta)}}}~\cos\theta ~d\theta\\=\int{\frac{1}{\sin^2\theta\cdot \cos\theta }}~\cos\theta d\theta \\=\int{\csc^2\theta }~d\theta \\=-\cot\theta \\=-\frac{\cos\theta }{\sin\theta }\\=-\frac{\sqrt{1-\sin^2\theta }}{\sin\theta }\\=-\frac{\sqrt{1-x^2}}{x}\rightarrow(2)$

Hence, by $~(1),~(2)~$ we get,

$~~I\\=-\frac{\cos^{-1}x}{2x^2}-\frac 12\left(-\frac{\sqrt{1-x^2}}{x}\right)\\=\frac{1}{2x^2}(x\sqrt{1-x^2}-\cos^{-1}x)+c.$

$29.~~\displaystyle\int{\tan^{-1}\sqrt{\frac{1-x}{1+x}}}~dx$

Solution.

$~\text{let}~x=\cos2\theta \ \therefore ~dx=-2\sin2\theta~d\theta.$

$I\\=\displaystyle\int{\tan^{-1}\sqrt{\frac{1-x}{1+x}}}~dx\\=\int{\tan^{-1}{\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}}(-2\sin2\theta)}~d\theta\\=-2\int{\tan^{-1}{\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}} \cdot \sin2\theta}~d\theta\\=-2\int{\tan^{-1}(\tan\theta) \cdot \sin2\theta}\\=-2\int{\theta \sin(2\theta)}~d\theta\\=-2I_1.$

$I_1\\=\int{\theta\sin2\theta}~d\theta\\=\theta\int{\sin2\theta}~d\theta-\int{\left[\frac{d}{d\theta}(\theta)\int{\sin2\theta}~d\theta\right]}~d\theta\\=\theta \cdot \left(-\frac{\cos2\theta}{2}\right)-\int{\left(-\frac{\cos2\theta}{2}\right) }\theta\\=-\frac 12(\theta\cos2\theta)+\frac 12\int{\cos2\theta}~d\theta\\=-\frac 12(\theta\cos2\theta)+\frac 12 \cdot \frac{\sin2\theta}{2}$

$I\\=-2\left[-\frac 12(\theta\cos2\theta)+\frac 12 \cdot \frac{\sin2\theta}{2}\right]\\=\theta\cos2\theta-\frac 12\sin2\theta+c\\=\theta\cos2\theta-\frac 12\sqrt{1-\cos^22\theta}+c\\=\frac 12\cos^{-1}x \cdot x-\frac{\sqrt{1-x^2}}{2}+c\\=\frac 12(x\cos^{-1}x-\sqrt{1-x^2}) +c.$

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Short Answer Type Questions of Integration By Parts :

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