Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Integrate the function $x \sin^{-1} x$
✓
Answer
Let $I = x \sin^{-1 }x$
Now, integrating by parts, we get,
$I=\sin ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right\} d x $
$= \sin ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x $
$= \frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x $
$= \frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\frac{1-x^{2}}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x $
$= \frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\sqrt{1-x^{2}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x $
$= \frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\left\{\int \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}} d x\right\} $
$= \frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C $
$= \frac{1}{4}\left(2 x^{2}-1\right) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^{2}}+C$
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