JEE Main 12 Jan 2019 Morning Question 24
Question: Integral $ \int _{{}}^{{}}{\cos (log _{e}x)}dx $ equals (where C is the constant of integration) [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{x}{2}[sin(log _{e}x)-cos(log _{e}x)]+C $
B) $ x[cos(log _{e}x)-\sin (log _{e}x)]+C $
C) $ \frac{x}{2}[cos(log _{e}x)+\sin (log _{e}x)]+C $
D) $ x[cos(log _{e}x)+\sin (log _{e}x)]+C $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ I=\int _{{}}^{{}}{\cos (log _{e}x)}dx $
$ \Rightarrow $ $ I=\cos (log _{e}x)x+\int _{{}}^{{}}{sin(log _{e}x)}dx $
$ \Rightarrow $ $ I=x\cos ({\log_e}x)+x\sin ({\log_e}x)x-\int _{{}}^{{}}{\cos ({\log_e}x)}dx $
$ \Rightarrow $ $ I=\frac{x}{2}[\cos ({\log_e}x)+\sin ({\log_e}x)]+C $
