First slide

Evaluation of definite integrals

Question

$The value of \int_{0}^{\sin^{2} x} \sin^{- 1} \sqrt{t} d t + \int_{0}^{\cos^{2} x} \cos^{- 1} \sqrt{t} d t is$

Moderate

A

$π$
B

$\frac{π}{2}$
C

$\frac{π}{4}$
D

1

Solution

$\begin{matrix} {Let I}_{1} = \int_{0}^{\sin^{2} x} \sin^{- 1} \sqrt{t} dt \\ Put t = \sin^{2} u \Rightarrow dt = 2 \sin u \cos udu \\ \Rightarrow dt = \sin 2 udu \\ ∴ I_{1} = \int_{0}^{x} u \sin 2 udu \\ {Let I}_{2} = \int_{0}^{\cos^{2} x} \cos^{- 1} \sqrt{t} d \\ Put t = \cos^{2} v \Rightarrow d t = - 2 \cos v \sin v d v \\ \Rightarrow dt = - \sin 2 vdv \\ ∴ I_{2} = \int_{\frac{π}{2}}^{x} v (- \sin 2 v) d v = - \int_{\frac{π}{2}}^{x} v \sin 2 v d v \\ = - \int_{\frac{π}{2}}^{x} u \sin 2 udu [change of variable] \\ ∴ I = I_{1} + I_{2} = \int_{0}^{x} u \sin 2 udu - \int_{\frac{π}{2}}^{x} u \sin 2 udu \\ = \int_{0}^{\frac{π}{2}} u \sin 2 udu + \int_{\frac{π}{2}}^{x} u \sin 2 udu - \int_{\frac{π}{2}}^{x} u \sin 2 udu \\ = \int_{0}^{\frac{π}{2}} u \sin 2 udu = \frac{π}{4} [Integratebyparts] \end{matrix}$

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