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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Integrate the function in exercise.
$\text{x}\ \sin^{-1}\text{x dx}$
✓

Answer

Let $\text{I}=\int\text{x}\sin^{-1}\text{x dx}$
Taking $\sin^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain.
$\text{I}=\sin^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$
$=\sin^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}\text{dx}$
$=\sin^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}+\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\frac{-\text{x}^2}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\Bigg\{\frac{1-\text{x}^2}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{x}^2}}\Bigg\}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\Bigg\{\sqrt{1-\text{x}^2}-\frac{1}{\sqrt{1-\text{x}^2}}\Bigg\}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\Bigg\{\int\sqrt{1-\text{x}^2}\text{dx}-\int\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\Bigg\{\frac{\text{x}}{2}\sqrt{1-\text{x}^2}+\frac{1}{2}\sin^{-1}\text{x}-\sin^{-1}\text{x}\Bigg\}+\text{C}$
$=\frac{1}{4}(2\text{x}^2-1)\sin^{-1}\text{x}+\frac{\text{x}}{4}\sqrt{1-\text{x}^2}+\text{C}$

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