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  • Integrate the functions in Exercises 1 to 24. 12

Integrate the functions in Exercises 1 to 24.

    Q12.    \frac{x^3}{\sqrt{1-x^8}}

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Given that to integrate

\frac{x^3}{\sqrt{1-x^8}}

Let x^4 = t \implies 4x^3dx = dt

\therefore \int \frac{x^3}{\sqrt{1-x^8}}dx = \frac{1}{4}\int\frac{1}{\sqrt {1-t^2}}dt

= \frac{1}{4}sin^{-1}t + C= \frac{1}{4}sin^{-1}{x^4} + C

the required solution is  \frac{1}{4}sin^{-1}{(x^4)} + C

Posted by

HARSH KANKARIA

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