Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 39 Maths Textbook Solution.

Answer: $\frac{-\sin ^{-1}x}{x}+\frac{1}{2}\left [ log\frac{\sqrt{1-x^{2}-1}}{1+x^{2}+1} \right ]+c$

Given: $\int \frac{sin^{-1}x}{x^{2}}dx$

Hint: $\int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx$

Solution:

\begin{aligned} &I=\int \frac{\sin ^{-1} x}{x^{2}} d x \\ &\int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x \\ &I=\sin ^{-1} \int \frac{1}{x^{2}} d x-\left[d \frac{\sin ^{-1} x}{d x}\left(\int \frac{1}{x^{2}} d x\right) d x\right] \\ &I=\frac{-\sin ^{-1} x}{x}+\int \frac{d x}{x \sqrt{1-x^{2}}} \end{aligned}

Let

$\sqrt{1-x^{2}}=u$

$\Rightarrow 1-x^{2}=u^{2}$

$\Rightarrow \frac{-2x}{2\sqrt{1-x^{2}}}dx=du$

$\Rightarrow \frac{-xdx}{\sqrt{1-x^{2}}}=du$

$\therefore -du=\frac{xdx}{\sqrt{1-x^{2}}}$

Now,

\begin{aligned} &y=\int \frac{d x}{x \sqrt{1-x^{2}}} \\ &\Rightarrow \int \frac{x d x}{x^{2} \sqrt{1-x^{2}}} \\ &\Rightarrow-\int \frac{d u}{1-u^{2}} \\ &\Rightarrow \frac{d u}{u^{2}-1} \\ &\Rightarrow \int \frac{d u}{(u+1)(u-1)} \Rightarrow \frac{1}{2} \int \frac{2 d u}{(u+1)(u-1)} \end{aligned}

\begin{aligned} &\Rightarrow \frac{1}{2} \int \frac{(u+1)-(u-1)}{(u+1)(u-1)} d u \\ &\Rightarrow \frac{1}{2}\left[\int \frac{(u+1)}{(u+1)(u-1)} d u-\int \frac{(u-1)}{(u+1)(u-1)} d u\right] \\ &\Rightarrow \frac{1}{2}[\log (u-1)-\log (u+1)] \\ &\Rightarrow \frac{1}{2}\left[\log \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}\right]+c \\ &I=\frac{-\sin ^{-1} x}{x}+\int \frac{d x}{x \sqrt{1-x^{2}}} \\ &=\frac{-\sin ^{-1} x}{x}+\frac{1}{2}\left[\log \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}\right]+c \end{aligned}