Integration By Parts (Part-8)| S N Dey| Class 12

by

In the previous article, we have discussed the solutions of 10 Long Answer type questions of Integration By Parts Chapter of S N Dey mathematics Class 12. In this chapter, we will discuss few more problems of Long Answer Type Questions of Exercise – 7 of Chhaya mathematics.

Integration By Parts, S N dey mathematics
Integration By Parts | Long Answer Type Questions (21-30) | S N Dey

21.~~ \displaystyle\int{\frac{\tan^{-1}x}{(1+x^2)^{3/2}}}~dx

Solution.

I=\displaystyle\int{\frac{\tan^{-1}x}{(1+x^2)^{3/2}}}~dx=\int{\frac{\tan^{-1}x}{(1+x^2)\sqrt{1+x^2}}}~dx \rightarrow(1)

\text{let}~\tan^{-1}x=\theta \\ \therefore~ \frac{dx}{1+x^2}=d\theta \rightarrow(2)

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

I\\=\displaystyle\int{\frac{\theta}{\sqrt{1+\tan^2\theta}}}~d\theta\\=\int{\frac{\theta}{\sec\theta}}~~d\theta\\=\int{\theta \cos\theta}~d\theta\\=\theta\int{\cos\theta}~d\theta- \int{\left[\frac{d}{d\theta}(\theta) \int{\cos\theta}~d\theta\right]}~d\theta\\=\theta \sin\theta-\int{\sin\theta}~d\theta\\=\theta\sin\theta+\cos\theta+c\\= \tan^{-1}x \cdot \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}+c\\=\frac{1}{\sqrt{1+x^2}}(x\tan^{-1}x+1)+c.

22.~~ \displaystyle\int{\frac{\log(x+1)}{\sqrt{x+1}}}~dx

Solution.

I=\displaystyle\int{\frac{\log(x+1)}{\sqrt{x+1}}}~dx\rightarrow(1)

~\text{let}~1+x=z^2 \\ \therefore dx=2z~dz \rightarrow(2)

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

I\\=\displaystyle \int{\frac{\log z^2}{z} \cdot 2z}~dz\\=4\int{\log z}~dz=4I_1 \rightarrow(3)

I_1\\=\displaystyle\int{\log z}~dz\\=\log z \int{dz}-\int{\left[\frac{d}{dz}(\log z) \int{}~dz\right]}~dz\\=z\log z- \int{ \frac 1z \cdot z}~dz\\=z\log z-\int{dz}\\=z\log z-z+c_1 \rightarrow(4)

Hence, by ~(3),~(4)~ we get,

I=4(z\log z-z+c_1)=4z \log z-4z+c.~~[~c=4c_1~]

23.~~\displaystyle\int{\sin^{-1}\sqrt{\frac{x}{a+x}}}~dx

Solution.

I=\displaystyle\int{\sin^{-1}\sqrt{\frac{x}{a+x}}}~dx \rightarrow(1)

\text{let}~~x=a\tan^2 \theta \\ \therefore dx=a \cdot 2\tan\theta \sec^2\theta~d\theta=2a\tan\theta\sec^2\theta d\theta \rightarrow(2)

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

I\\=\displaystyle\int{\sin^{-1}\sqrt{\frac{a\tan^2\theta }{a+a\tan^2\theta }}}~ 2a\tan\theta \sec^2\theta d\theta \\=2a\int{\sin^{-1}\sqrt{\frac{a\tan^2\theta }{a(1+\tan^2\theta )}}}\tan \theta \sec^2\theta d\theta \\=2a \int{\sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}}}~ \tan \theta \sec^2\theta d\theta\\=2a\int{\sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right)}\tan \theta \sec^2\theta d\theta\\=2a\int{\sin^{-1}\left(\frac{\sin\theta}{\cos\theta \cdot \sec\theta}\right)}\tan \theta \sec^2\theta d\theta\\=2a\int{\sin^{-1}(\sin\theta)}~\tan \theta \sec^2\theta d\theta\\=2a\int{\theta}\tan \theta \sec^2\theta d\theta\\=2aI_1\rightarrow(3)

~\text{let}~\tan\theta=z \\ \therefore~ \sec^2\theta d\theta=dz \rightarrow(4)

Hence, by ~(3),~(4)~ we get,

I_1\\=\displaystyle\int{\theta}\tan \theta \sec^2\theta d\theta\\=\int{z\tan^{-1}z}~dz\\=\tan^{-1}z \int{z~dz}-\int{\left[\frac{d}{dz}(\tan^{-1}z) \int{z}~dz\right]}~dz\\=\frac{z^2}{2} \cdot \tan^{-1}z-\int{\frac{1}{1+z^2} \cdot \frac{z^2}{2}}~dz\\=\frac{z^2}{2} \cdot \tan^{-1}z-\frac 12 \int{\left(1-\frac{1}{1+z^2}\right)}~dz\\=\frac{z^2}{2} \cdot \tan^{-1}z-\frac 12 \int{dz}-\frac 12\int{\frac{dz}{1+z^2}}\\=\frac{z^2}{2} \cdot \tan^{-1}z-\frac 12 z+\frac 12~\tan^{-1}z\\=\frac 12 \theta \tan^2\theta-\frac 12\tan\theta+\frac 12\theta\rightarrow(5)

Hence, by ~(3),~(5)~ we get,

I\\=2a \left[\frac 12 \theta \tan^2\theta-\frac 12\tan\theta+\frac 12\theta\right]+c\\=a\theta\tan^2\theta-a\tan\theta+a\theta+c\\=a\theta(\tan^2\theta+1)-a\tan\theta+c\\=a(\tan^{-1}\sqrt{x/a})~(1+x/a)-a\sqrt{\frac xa}+c\\=(x+a)\tan^{-1}\sqrt{x/a}-\sqrt{ax}+c.

24.~~\displaystyle\int{\frac{\sin(\log x)}{x^3}}~dx

Solution.

I=\displaystyle\int{\frac{\sin(\log x)}{x \cdot x^2}}~dx\rightarrow(1)

~\text{let}~~ \log x=z ~~(\Rightarrow x=e^z) \\ \therefore~ \frac{dx}{x}=dz\rightarrow(2)

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

I\\=\displaystyle\int{\frac{\sin z}{(e^z)^2}}~dz\\=\int{e^{-2z}\sin z}~dz\\=\frac{e^{-2z}}{(-2)^2+1^2}[-2\sin z-\cos z]+c\\=-\frac{e^{-2z}}{5}(2\sin z+\cos z)+c\\=-\frac{1}{5e^{2z}}(2\sin z+\cos z)+c\\=-\frac{1}{5x^2}[2\sin(\log x)+\cos(\log x)]+c.

25(i)~~\displaystyle\int{\sqrt{4x^2-9}}~dx

Solution.

I=~\displaystyle\int{\sqrt{4x^2-9}}~dx=\displaystyle\int{\sqrt{(2x)^2-3^2}}~dx\rightarrow(1)

~\text{let}~2x=z \Rightarrow dx=\frac 12dz \rightarrow(2)

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

I\\=\frac 12\displaystyle\int{\sqrt{z^2-3^2}}~dz\\=\frac 12 \left[\frac z2\sqrt{z^2-3^2}--\frac{3^2}{2}\log|z+\sqrt{z^2-3^2}|\right]+c_1\\=\frac 12\left[x\sqrt{4x^2-9}-\frac 92\log|2x+\sqrt{4x^2-9}|\right]+c_1\\=\frac 12\left[2x\sqrt{x^2-\frac 94}-\frac 92\log\left|2x+2\sqrt{x^2-\frac 94}\right| \right]+c_1\\=x\sqrt{x^2-\frac 94}-\frac 94 \log\left|2\left(x+\sqrt{x^2-\frac 94}\right)\right|+c_1\\=x\sqrt{x^2-\frac 94}-\frac 94\log 2-\frac 94 \log\left|x+\sqrt{x^2-\frac 94}\right|+c_1\\=x\sqrt{x^2-\frac 94}-\frac 94 \log\left|x+\sqrt{x^2-\frac 94}\right|+c. ~~\left[~c=c_1-\frac 94\log 2~\right]

25(ii)~~ \displaystyle\int{\sqrt{x(x^3+a^3)}}~dx

Solution.

I\\= \displaystyle\int{\sqrt{x(x^3+a^3)}}~dx\\= \displaystyle\int{\sqrt{x}\sqrt{(x^{3/2})^2+(a^{3/2})^2}}~dx \rightarrow(1)

\text{let}~~x^{3/2}=z \\ \therefore~ \frac 32 x^{\frac 32-1}~dx=dz \\ \text{or,}~~ \sqrt{x}~dx=\frac 23 ~dz\rightarrow(2)

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

I\\=\frac 23 \displaystyle\int{\sqrt{z^2+(a^{3/2})^2}}~dz\\=\frac 23\left[\frac{z\sqrt{z^2+(a^{3/2})^2}}{2}+\frac{(a^{3/2})^2}{2}\log\left|z+\sqrt{z^2+(a^{3/2})^2}\right|\right]+c\\=\frac{x^{3/2}}{3}\sqrt{x^3+a^3}+\frac{a^3}{3}\log|x^{3/2}+\sqrt{x^3+a^3}|+c.

26.~~ \displaystyle\int{\sqrt{4-9x^2}}~dx

Solution.

\displaystyle\int{\sqrt{4-9x^2}}~dx\\=3\int{\sqrt{\left(\frac 23\right)^2-x^2}}~dx\\=3\left[\frac x2 \sqrt{\left(\frac 23\right)^2-x^2}+\frac{(2/3)^2}{2}\sin^{-1} \frac{x}{(2/3)}\right]+c\\=3\left[\frac x2 \cdot \frac{\sqrt{4-9x^2}}{3}+\frac 29 \sin^{-1}\left(\frac{3x}{2}\right)\right]+c\\=\frac x2\sqrt{4-9x^2}+\frac 23\sin^{-1}\left(\frac{3x}{2}\right)+c.

27.~~ \displaystyle\int{\sqrt{x^2+x+1}}~dx

Solution.

\displaystyle\int{\sqrt{x^2+x+1}}~dx\\= \displaystyle\int{\sqrt{x^2+2 \cdot x \cdot \frac 12+\left(\frac 12 \right)^2+1-\left(\frac 12\right)^2}}~dx \\= \displaystyle\int{\sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}~dx\\= \displaystyle\int{\sqrt{z^2+(\sqrt{3}/2)^2}}~dz~~[~\text{let}~ z=x+1~] \\= \frac z2 \sqrt{z^2+(\sqrt{3}/2)^2}+\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2} \log\left|z+\sqrt{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c\\=\frac{x+\frac 12}{2} \sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac 38 \log \left|x+\frac 12+\sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c\\=\frac{2x+1}{4} \sqrt{x^2+x+1}+\frac 38 \log\left|x+\frac 12 +\sqrt{x^2+x+1}\right|+c.

28.~~ \displaystyle\int{\sqrt{2ax-x^2}}~dx

Solution.

\displaystyle\int{\sqrt{2ax-x^2}}~dx\\=\int{a^2-(x^2-2xa+a^2)}~dx\\=\int{\sqrt{a^2-(x-a)^2}}~dx \rightarrow(1)

\text{let}~x-a=z \Rightarrow dx=dz \rightarrow(2)

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

\displaystyle\int{\sqrt{2ax-x^2}}~dx\\=\int{\sqrt{a^2-z^2}}~dz\\=\frac z2 \sqrt{a^2-z^2}+\frac{a^2}{2}\sin^{-1}(z/a)+c\\=\frac{x-a}{2}~\sqrt{2ax-x^2}+\frac{a^2}{2} \sin^{-1}\left(\frac{x-a}{a}\right)+c.

29.~~ \displaystyle\int{\sqrt{4-3x-2x^2}}~dx

Solution.

\displaystyle\int{\sqrt{4-3x-2x^2}}~dx\\=\sqrt{2} \int{\sqrt{2-\frac 32 x-x^2}}~dx\\= \sqrt{2}\int{\sqrt{-\left(x^2+\frac 32 x\right)+2}}~dx\\= \sqrt{2}\int{\sqrt{2-\left[x^2+2 \cdot x \cdot \frac 34+(3/4)^2\right]+\left(\frac{3}{4}\right)^2}}~dx\\= \sqrt{2}\int{\sqrt{\left(2+\frac{9}{16}\right)-(x+3/4)^2}}~dx\\=\sqrt{2}\int{\sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x+\frac 34\right)^2}}~dx\\= \sqrt{2}\left[\frac{x+\frac 34}{2} \sqrt{\left(\frac{\sqrt{41}}{4}\right)^2-\left(x+\frac 34\right)^2}+\frac{\left(\sqrt{41}/4\right)^2}{2}\sin^{-1} \frac{x+\frac 34}{\frac{\sqrt{41}}{4}}\right]+c\\=\sqrt{2}\left[\frac{4x+3}{8} \sqrt{\frac{41}{16}-\frac{(4x+3)^2}{16}}+\frac{41}{32} \sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right)\right]+c\\=\sqrt{2}\left[\frac{4x+3}{8} \sqrt{\frac{41-(16x^2+24x+9)}{16}}+\frac{41}{32} \sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right)\right]+c\\=\sqrt{2}\left[\frac{4x+3}{8} \sqrt{\frac{8(4-2x^2-3x)}{16}}+\frac{41}{32} \sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right)\right]+c\\=\sqrt{2}\left[\frac{4x+3}{8} \cdot \frac{1}{\sqrt{2}}\sqrt{4-3x-2x^2}+\frac{41}{32} \sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right)\right]+c\\=\frac{4x+3}{8}\sqrt{4-3x-2x^2}+\frac{41\sqrt{2}}{32} \sin^{-1}\left(\frac{4x+3}{\sqrt{41}}\right)+c

30.~~ \displaystyle\int{\sqrt{(x-2)(6-x)}}~dx

Solution.

\displaystyle\int{\sqrt{(x-2)(6-x)}}~dx\\=\displaystyle\int{\sqrt{-12+8x-x^2}}~dx\\=\displaystyle\int{\sqrt{-(x^2-8x+12)}}~dx\\=\displaystyle\int{\sqrt{-(x^2-2 \cdot x \cdot 4+4^2)+16-12}}~dx\\=\displaystyle\int{\sqrt{4-(x-4)^2}}~dx\\=\displaystyle\int{\sqrt{2^2-(x-4)^2}}~dx\\=\frac{x-4}{2} \sqrt{2^2-(x-4)^2}+\frac{2^2}{2}\sin^{-1}\left(\frac{x-4}{2}\right)+c\\=\frac{x-4}{2} \sqrt{(x-2)(6-x)}+2\sin^{-1}\left(\frac{x-4}{2}\right)+c\\=\frac 12 \left[(x-4)\sqrt{(x-2)(6-x)}+4\sin^{-1}\left(\frac{x-4}{2}\right)\right]+c.

Post Views: 123

Leave a Comment