Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$
✓
Answer
Let first function be $\sin^{-1} x$ and second function be $\frac{x}{\sqrt{1-x^{2}}}$.
First we find the integral of the second function, i.e., $\int \frac{x d x}{\sqrt{1-x^{2}}}$
Put $t =1 - x^2$. Then dt = -2x dx
Therefore, $\int \frac{x d x}{\sqrt{1-x^{2}}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^{2}}$
Hence, $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$ = $(\sin ^{-1} x)$$\int \frac{x d x}{\sqrt{1-x^{2}}}$ $-\int [\frac d{dx} sin^{-1}x.\int \frac{x d x}{\sqrt{1-x^{2}}} ] d x$
= $\left(\sin ^{-1} x\right)(-\sqrt{1-x^{2}})-\int \frac{1}{\sqrt{1-x^{2}}}(-\sqrt{1-x^{2}}) d x$
= $-\sqrt{1-x^{2}} \sin ^{-1} x+x+C$
= $x-\sqrt{1-x^{2}} \sin ^{-1} x+\mathrm{C}$
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