A) \[\int{{{e}^{\sin \theta }}[\log \sin \theta +\cos e{{c}^{2}}\theta ]}+c\]
B) \[{{e}^{\sin \theta }}[\log \sin \theta +\cos ec\theta ]+c\]
C) \[{{e}^{\sin \theta }}[\log \sin \theta -\cos ec\theta ]+c\]
D) \[{{e}^{\sin \theta }}[\log \sin \theta -\cos e{{c}^{2}}\theta ]+c\]
E) \[{{e}^{\sin \theta }}[\log \sin \theta +{{\cos }^{2}}\theta ]+c\]
Correct Answer: C
Solution :
Let \[I=\int{{{e}^{\sin \theta }}\log \sin \theta \cos \theta d\theta }\] \[+\int{{{e}^{\sin \theta }}\cos e{{c}^{2}}\theta \cos \theta d\theta }\] Put\[\sin \theta =t\Rightarrow \cos \theta d\theta =dt\] \[\therefore \]\[I=\int{{{e}^{t}}\log t\,dt+\int{\frac{{{e}^{t}}}{t}dt}}+\frac{{{e}^{t}}{{t}^{-1}}}{-1}-\int{\frac{{{e}^{t}}{{t}^{-1}}}{-1}}dt\] \[={{e}^{t}}\left( \log t-\frac{1}{t} \right)+c\] \[={{e}^{\sin \theta }}(\log \sin \theta -\cos ec\theta )+c\]
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