A) \[-\frac{1}{2}\cos 4x+c\]
B) \[-\frac{1}{4}\cos 4x+c\]
C) \[-\frac{1}{2}\sin 2x+c\]
D) none of these
Correct Answer: D
Solution :
Let \[I=\int_{{}}^{{}}{\frac{\cos 4x-1}{\cot x-\tan x}}dx\] \[=-\int_{{}}^{{}}{\frac{2{{\sin }^{2}}2x}{\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\cos x\sin x}}dx}\] \[=-\int_{{}}^{{}}{\frac{\sin 2x(1-co{{s}^{2}}2x)}{\cos 2x}}dx\] Put \[\cos 2x=t\]\[\Rightarrow \]\[-2\sin 2x\,dx=dt\] \[\therefore \] \[I=\frac{1}{2}\int_{{}}^{{}}{\frac{(1-{{t}^{2}})}{t}}dt\] \[=\frac{1}{2}\left[ \int_{{}}^{{}}{\left( \frac{1}{2}-t \right)dt} \right]\] \[=\frac{1}{2}\left[ \log t-\frac{{{t}^{2}}}{2} \right]+c\] \[=\frac{1}{2}\log \cos 2x-\frac{{{\cos }^{2}}2x}{4}+c\]
You need to login to perform this action.
You will be redirected in
3 sec
