First slide

Evaluation of definite integrals

Question

$\int_{0}^{\sin^{2} x} \sin^{- 1} \sqrt{t} d t + \int_{0}^{\cos^{2} x} \cos^{- 1} \sqrt{t} d t =$

Difficult

A

$\frac{π}{4}$
B

$\frac{π}{3}$
C

$\frac{π}{2}$
D

$π$

Solution

Let $t = \sin^{2} u$ in the first integral and $t = \cos^{2} u$ in the second integral
$I_{1} + I_{2} = \int_{0}^{x} u \sin 2 u d u + \int_{x}^{\frac{π}{2}} u \sin 2 u d u = \int_{0}^{\frac{π}{2}} u \sin 2 u d u$

$= - \frac{u}{2} \cos 2 u {+ \frac{1}{4} \sin 2 u |}_{0}^{\frac{π}{2}} = \frac{π}{4}$

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