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Question: \(\int _ { \pi / 4 } ^ { \pi / 2 } \cos \theta \operatorname { cosec } ^ { 2 } \theta d \theta =\)...

∫π/4π/2cos⁡θcosec⁡2θdθ=\int _ { \pi / 4 } ^ { \pi / 2 } \cos \theta \operatorname { cosec } ^ { 2 } \theta d \theta =

A

2−1\sqrt { 2 } - 1

B

1−21 - \sqrt { 2 }

C

2+1\sqrt { 2 } + 1

D

None of these

Answer

2−1\sqrt { 2 } - 1

Explanation

Solution

Let ∫π/4π/2cos⁡θ1sin⁡2θdθ\int _ { \pi / 4 } ^ { \pi / 2 } \cos \theta \frac { 1 } { \sin ^ { 2 } \theta } d \theta

Put t=sin⁡θ⇒dt=cos⁡θdθt = \sin \theta \Rightarrow d t = \cos \theta d \theta then we have

∫1/211t2dt=[−1t]1/21=2−1\int _ { 1 / \sqrt { 2 } } ^ { 1 } \frac { 1 } { t ^ { 2 } } d t = \left[ \frac { - 1 } { t } \right] _ { 1 / \sqrt { 2 } } ^ { 1 } = \sqrt { 2 } - 1.