A) \[{{\cos }^{2}}x\left( \frac{1}{2}+\log \cos x \right)+k\]
B) \[{{\cos }^{2}}x.\log \,\,\cos \,\,x+k\]
C) \[{{\cos }^{2}}x\left( \frac{1}{2}-\log \cos x \right)+k\]
D) None of these
Correct Answer: C
Solution :
[c] \[I=\int{2\sin x.\cos x.\log \,\,\cos \,\,x\,\,dx}\] Put \[\log \,\,\cos \,\,x=t\] \[\therefore -\frac{\sin \,\,x}{\cos \,\,x}dx=dt\] \[I=\int{2\sin x.\cos x.t\frac{\cos x}{-\sin x}dt}\] \[=-2\int{{{\cos }^{2}}x.t\,\,dt=-2\int{t{{e}^{2t}}dt}}\] \[=-2\left[ t.\frac{{{e}^{2t}}}{2}-\int{\frac{{{e}^{2t}}}{2}.dt} \right]=-t{{e}^{2t}}+\frac{1}{2}{{e}^{2t}}+k\] \[{{e}^{2t}}\left( \frac{1}{2}-t \right)+k={{\cos }^{2}}x.\left\{ \frac{1}{2}-\log \cos x \right\}+k\]
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