Question
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}$
$\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}\ ....(1)$ Let $\sin^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\sin^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow2\text{x}\times\frac{1}{\sqrt{1-\text{x}^4}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{x}}{\sqrt{1-\text{x}^4}}\text{ dx}=\frac{\text{dt}}{2}$ Putting $\sin^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{\sqrt{1-\text{x}}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),
We get,
$\text{I}=\int\text{t}\frac{\text{dt}}{2}$ $=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$ $=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$ $\text{I}=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$Need a full question paper?
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