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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
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Answer

Let $\text{I}=\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
Let $\sin^{-1}\text{x}^2=\text{t}$
$\frac{1}{\sqrt{1-\text{x}^4}}(2\text{x})\text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^2}}\text{x dx}$
$=\int(\sin\text{t})\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\int\text{t}\sin\text{t dt}$
$=\frac{1}{2}\big[\text{t}\int\sin\text{t dt}-\int(1\int\sin\text{t dt})\text{dt}\big]$
$=\frac{1}{2}\big[\text{t}(-\cos\text{t})-\int(-\cos\text{t})\text{dt}\big]$
$=\frac{1}{2}\big[-\text{t}\cos\text{t}+\sin\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{x}^2-\sqrt{1-\text{x}^4}\sin^{-1}\text{x}^2\Big]+\text{C}$

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