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Question

Question: The value of \(\int _ { 0 } ^ { \sin ^ { 2 } x } \sin ^ { - 1 } \sqrt { t } d t + \int _ { 0 } ^ { ...

The value of 0sin2xsin1tdt+0cos2xcos1tdt\int _ { 0 } ^ { \sin ^ { 2 } x } \sin ^ { - 1 } \sqrt { t } d t + \int _ { 0 } ^ { \cos ^ { 2 } x } \cos ^ { - 1 } \sqrt { t } d t is

A

π2\frac { \pi } { 2 }

B

1

C

π4\frac { \pi } { 4 }

D

None of these

Answer

π4\frac { \pi } { 4 }

Explanation

Solution

We have

I=0sin2xsin1tdt+0cos2xcos1tdtI = \int _ { 0 } ^ { \sin ^ { 2 } x } \sin ^ { - 1 } \sqrt { t } d t + \int _ { 0 } ^ { \cos ^ { 2 } x } \cos ^ { - 1 } \sqrt { t } d t

Putting t=sin2ut = \sin ^ { 2 } uin the first integral and t=cos2vt = \cos ^ { 2 } v in the second integral, we have

I=0xusin2uduπ/2xvsin2vdvI = \int _ { 0 } ^ { x } u \sin 2 u d u - \int _ { \pi / 2 } ^ { x } v \sin 2 v d v

=0π/2usin2udu+π/2xusin2uduπ/2xvsin2vdv= \int _ { 0 } ^ { \pi / 2 } u \sin 2 u d u + \int _ { \pi / 2 } ^ { x } u \sin 2 u d u - \int _ { \pi / 2 } ^ { x } v \sin 2 v d v

I=0π/2usin2udu=(ucos2u2)0π/2+120π/2cos2uduI = \int _ { 0 } ^ { \pi / 2 } u \sin 2 u d u = \left( \frac { - u \cos 2 u } { 2 } \right) _ { 0 } ^ { \pi / 2 } + \frac { 1 } { 2 } \int _ { 0 } ^ { \pi / 2 } \cos 2 u d u

=(ucos2u2)0π/2+14(sin2u)0π/2=π4= \left( \frac { - u \cos 2 u } { 2 } \right) _ { 0 } ^ { \pi / 2 } + \frac { 1 } { 4 } ( \sin 2 u ) _ { 0 } ^ { \pi / 2 } = \frac { \pi } { 4 } .