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Question: \(\int _ { 0 } ^ { \pi / 2 } \sqrt { \cos \theta } \sin ^ { 3 } \theta d \theta =\)...

∫0π/2cos⁡θsin⁡3θdθ=\int _ { 0 } ^ { \pi / 2 } \sqrt { \cos \theta } \sin ^ { 3 } \theta d \theta =

A

2021\frac { 20 } { 21 }

B

821\frac { 8 } { 21 }

C

−2021\frac { - 20 } { 21 }

D

−821\frac { - 8 } { 21 }

Answer

821\frac { 8 } { 21 }

Explanation

Solution

LetI=∫0π/2cos⁡θsin⁡3θdθI = \int _ { 0 } ^ { \pi / 2 } \sqrt { \cos \theta } \sin ^ { 3 } \theta d \theta

Put t=cos⁡θ⇒dt=−sin⁡θdθt = \cos \theta \Rightarrow d t = - \sin \theta d \theta then

I =−∫10t1/2(1−t2)dt=∫01(t1/2−t5/2)- \int _ { 1 } ^ { 0 } t ^ { 1 / 2 } \left( 1 - t ^ { 2 } \right) d t = \int _ { 0 } ^ { 1 } \left( t ^ { 1 / 2 } - t ^ { 5 / 2 } \right) dtd t

I = [23t3/2−27t7/2]01=821\left[ \frac { 2 } { 3 } t ^ { 3 / 2 } - \frac { 2 } { 7 } t ^ { 7 / 2 } \right] _ { 0 } ^ { 1 } = \frac { 8 } { 21 }.