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

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

Answers (1)

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}

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