Integral Calculus Question 99
Question: $ \int_{{}}^{{}}{\sin (\log x)dx=} $
Options:
A) $ \frac{1}{2}x[\cos (\log x)-\sin (\log x)] $
B) $ \cos (\log x)-x $
C) $ \frac{1}{2}x[\sin (\log x)-\cos (\log x)] $
D) $ -\cos \log x $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ I=\int_{{}}^{{}}{\sin (\log x),dx} $ Put $ \log x=t\Rightarrow x=e^{t}\Rightarrow dx=e^{t}dt, $ then $ I=\int_{{}}^{{}}{\sin t,.,e^{t},dt}=\sin t,.,e^{t}-\int_{{}}^{{}}{e^{t}.,\cos t,dt} $ $ =\sin t,.,e^{t}-[\cos t,.,e^{t}+\int_{{}}^{{}}{e^{t}.\sin t,dt}] $
$ \Rightarrow 2I=\sin t,.,e^{t}-\cos t,.,e^{t} $
$ \Rightarrow I=\int_{{}}^{{}}{\sin ,(\log x),dx} $ $ =\frac{1}{2}x,[\sin ,(\log x)-\cos ,(\log x)] $ .
BETA
