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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:

$\int\text{x}^2\sin^{-1}\text{x dx}$

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Answer

Let $\text{I}=\int\text{x}^2\sin^{-1}\text{x dx}$
$\text{I}=\sin^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{\sqrt{1-\text{x}^2}}\int\text{x}^2\text{dx}\Big)\text{dx}$
$=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\int\frac{\text{x}^3}{3\sqrt{1-\text{x}^2}}\text{dx}$
$\text{I}=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\frac{1}{3}\text{I}_1+\text{C}_1\dots(1)$
$\text{I}_1=\int\frac{\text{x}^3}{\sqrt{1-\text{x}^2}}\text{dx}$
Let $1-\text{x}^2=\text{t}^2$
$-2\text{x dx}=2\text{t dt}$
$-\text{x dx}=\text{t dt}$
$\text{I}_1=-\int\frac{(1-\text{t}^2)\text{tdt}}{\text{t}}$
$=\int(\text{t}^2-1)\text{dt}$
$=\frac{\text{t}^3}{3}-\text{t}+\text{C}_2$
$=\frac{(1-\text{x}^2)^{\frac{3}{2}}}{3}-(1-\text{x}^2)^{\frac{1}{2}}+\text{C}_2$
Now,
$\text{I}=\frac{\text{x}^3}{3}\sin^{-1}\text{x}-\frac{1}{9}(1-\text{x}^2)^{\frac{3}{2}}+\frac{1}{3}(1-\text{x}^2)^{\frac{1}{2}}+\text{C}$

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