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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Integrate the function in Exercise:
$\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
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Answer

Let $\text{I}=\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$ $\text{I}=\frac{-1}{2}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}.\cos^{-1}\text{x}\text{dx}$ Taking $\cos^{-1}$ x as first function and $\Bigg(\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\Bigg)$ as second function and integrating by parts, we obtain. $\text{I}=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}\Bigg)\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}\text{dx}\Bigg]$$=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]$
$=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\int2\text{dx}\Big]$ $=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+2\text{x}\Big]+\text{C}$ $=-\Big[\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+\text{x}\Big]+\text{C}$

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