First slide

Evaluation of definite integrals

Question

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

Moderate

A

1
B

$\frac{π}{2}$
C

$\frac{π}{2} - 1$
D

$\frac{π}{2} + 1$

Solution

$I = 2 \int_{0}^{\frac{π}{2}} \sin x \cos x \tan^{- 1} (\sin x) d x, t = \sin x$
$= 2 \int_{0}^{1} t \tan^{- 1} t d t = 2 [{\frac{t^{2}}{2} \tan^{- 1} t |}_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{t^{2}}{t^{2} + 1} d t]$

$= \frac{π}{4} - \int_{0}^{1} \frac{t^{2} + 1 - 1}{t^{2} + 1} d t$

$= \frac{π}{4} - 1 + \frac{π}{4} = \frac{π}{2} - 1$

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