e^ cot ⁡x sin^2 ⁡x (2 log ⁡ cosec ⁡x+ sin ⁡2x ) dx=

Question

Accepted Answer

Correct answer: B. (b) e^ cot ⁡x( log ⁡ cosec^2 ⁡x+ 2 tan ⁡x1+ tan^2 ⁡x) cosec^2 ⁡ xdx Let cot ⁡x=t cosec^2 ⁡ xdx=- dt and tan ⁡x= 1t cosec^2 ⁡x=1+ cot^2 ⁡x=1+t^2=- e^t( log ⁡(1+t^2)+ 2 1t1+ 1t^2) dt =- e^t( log ⁡(1+t^2)+ 2t1+t^2)dt Since Integral is of the form - e^t(f (t )+f^ ' (t )) dt=-e^tf (t )+C ∵t= cot ⁡x=-e^ cot ⁡x log ⁡(1+ cot^2 ⁡x)+C=-e^ cot ⁡x ( log ⁡ cosec^2 ⁡x )+C=-2e^ cotr ( log ⁡ cosec^x )+C

$\sum \frac{e ^{cot x}}{sin ^{2} x} (2 log cosec x + sin 2 x) dx =$

Solution

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