If $I=\int e^{\sin heta}\left(\log \sin heta+\operatorname{cosec}^2 heta ight) \cos heta d heta$, then $I$ is equal to

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If $I=\int e^{\sin 	heta}\left(\log \sin 	heta+\operatorname{cosec}^2 	hetaight) \cos 	heta d 	heta$, then $I$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} 	herefore \quad & 	ext { Let } \sin 	heta=t \ 	herefore \quad & \cos 	heta d 	heta=d t \ 	herefore \quad & =\int e^t\left(\log t+\frac{1}{t^2}ight) d t \ & =\int e^t \log t d t+\int \frac{e^t}{t^2} d t \ & =\log t e^t-\int \frac{e^t}{t} d t+\left[\frac{e^t}{-t}-\int \frac{e^t}{-t} d tight]+c \ & =\log t e^t-\int \frac{e^t}{t} d t-\frac{e^t}{t}+\int \frac{e^t}{t} d t+c \ & =e^t\left[\log t-\frac{1}{t}ight]+c \ & =e^{\sin 	heta}[\log \sin 	heta-\operatorname{cosec} 	heta]+c\end{aligned}$

If $I=\int e^{\sin \theta}\left(\log \sin \theta+\operatorname{cosec}^2 \theta\right) \cos \theta d \theta$,…

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