Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Integrate the function in Exercise:
$(\sin^{-1}\text{x})^2$
✓
Answer
Let $\text{I}=\int(\sin^{-1}\text{x})^21.\text{dx}$
Taking $(\sin^{-1} x)^2$ as first function and 1 as second function and integrating by parts, we obtain.
$\text{I}=(\sin^{-1}\text{x})^2\int1\text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2\int1.\text{dx}\Bigg\}\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2-\int\frac{2\sin^{-1}\text{x }}{\sqrt{1-\text{x}^2}}.\text{x} \ \text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2.\int\sin^{-1}\text{x}.\Bigg(\frac{-2\text{x }}{\sqrt{1-\text{x}^2}}\Bigg)\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2+\Bigg[\sin^{-1}\text{x}\int\frac{-2\text{x }}{\sqrt{1-\text{x}^2}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}\Bigg)\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}\text{dx}\Bigg]$
$=\text{x}(\sin^{-1}\text{x})^2+\Bigg[\sin^{-1}\text{x}-2.\sqrt{1-\text{x}^2}-\int\frac{1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]$
$=\text{x}(\sin^{-1}\text{x})^2+2\sqrt{1-\text{x}^2}\sin^{-1}\text{x}-\int2\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2+2\sqrt{1-\text{x}^2}\sin^{-1}\text{x}-2\text{x}+\text{C}$
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