$\int_{}^{}{e^{\sin heta}\lbrack\log{\sin heta + cosec^{2} heta brack\cos heta}}d heta$ is equal to

Question

$\int_{}^{}{e^{\sin	heta}\lbrack\log{\sin	heta + cosec^{2}	hetabrack\cos	heta}}d	heta$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Let $I = \int_{}^{}{e^{\sin	heta}(\log{\sin	heta)\cos	heta}}d	heta + \int_{}^{}e^{\sin	heta}cosec^{2}	heta\cos	heta\ d	heta$
Put $\sin	heta = t\ \  \Rightarrow \cos	heta\ d	heta = dt$
$	herefore\ I = \int_{}^{}e^{t}\log t\ dt + \int_{}^{}{e^{t}\ t^{- 2}}dt$
$= \log t\ e^{t} - \int_{}^{}\frac{e^{t}}{t}\ dt + \frac{e^{t}t^{- 1}}{- 1} - \int_{}^{}\frac{e^{t}t^{- 1}}{- 1}\ dt$
$= e^{t}\left( \log t - \frac{1}{t} ight) + c = e^{\sin	heta}\left( \log{\sin	heta - cosec\ 	heta} ight) + c$

Q. $\int_{}^{}{e^{\sin\theta}\lbrack\log{\sin\theta + cosec^{2}\theta\rbrack\cos\theta}}d\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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