Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
∫e sin θ [ log sin θ + cos ec2θ ] cos θ dθ is equal to
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. $ \int{{{e}^{\sin \theta }}[\log \sin \theta +\cos e{{c}^{2}}\theta ]}\cos \,\theta d\theta $ is equal to
KEAM
KEAM 2008
Integrals
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 $
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 $