First slide

Introduction to definite integration

Question

$\int_{0}^{1} \frac{4 x^{3}}{\sqrt{1 - x^{8}}} d x =$

Very Easy

A

$π$
B

$- π$
C

$π / 2$
D

$- π / 2$

Solution

Put $x^{4} = t$ then the given integration becomes
$= \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = {[\sin^{- 1} t]}_{0}^{1} = \frac{π}{2}$

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