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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}$
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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,

$\text{I}=\int\text{t}\frac{\text{dt}}{2}$

$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$

$=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$

$\text{I}=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$

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