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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration3 Marks
Question
Evaluate : $\int x \cdot \sin ^{-1} x \cdot d x$
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Answer

$\mathrm{I}=\int \sin ^{-1} x \cdot x \cdot d x \quad \ldots$. by LIATE
$
\begin{aligned}
& =\sin ^{-1} x \cdot \int x \cdot d x-\int \frac{d}{d x} \cdot \sin ^{-1} x \cdot \int x \cdot d x \cdot d x \\
& =\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{1-\left(1-x^2\right)}{\sqrt{1-x^2}} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int\left[\frac{1}{\sqrt{1-x^2}}-\frac{\left(1-x^2\right)}{\sqrt{1-x^2}}\right] \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \int \frac{d x}{\sqrt{1-x^2}}+\frac{1}{2} \int \sqrt{1-x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x-\frac{1}{2} \sin { }^{-1} x+\frac{1}{2}\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)\right]+c \\
& =\frac{1}{2} x^2 \cdot \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1} x+c \\
\therefore \quad \int x \cdot \sin ^{-1} x \cdot d x & =\frac{1}{2} x^2 \cdot \sin ^{-1} x+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1} x+c
\end{aligned}
$

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