First slide

Evaluation of definite integrals

Question

$\int_{0}^{1} x \sin^{- 1} x d x =$

Easy

A

$\frac{π}{8}$
B

$\frac{π}{4}$
C

$π$
D

$\frac{π}{2}$

Solution

$\int_{0}^{1} x S i n^{- 1} x d x = {(S i n^{- 1} x . \frac{x^{2}}{2})}_{0}^{1} - \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} . \frac{x^{2}}{2} d x$
                        $= (\frac{π}{4} - 0) - \frac{1}{2} \int_{0}^{1} \frac{1 - (1 - x^{2})}{\sqrt{1 - x^{2}}} d x$

                         $= \frac{π}{4} - \frac{1}{2} \int_{0}^{1} (\frac{1}{\sqrt{1 - x^{2}}} - \sqrt{1 - x^{2}}) d x$

                         $= \frac{π}{4} - \frac{1}{2} {(S i n^{- 1} x)}_{0}^{1} + \frac{1}{2} \times \frac{π \times 1^{2}}{4}$

                          $= \frac{π}{8} (∵ \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x = \frac{π a^{2}}{4})$

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