Question

$\int \sin 2x.\log \cos xdx$ is equal to

• A. ${\cos }^{2}x\left(\frac{1}{2}+\log \cos x\right)+k$
• B. ${\cos }^{2}x\cdot \log \cos x+k$
• C. ${\cos }^{2}x\left(\frac{1}{2}-\log \cos x\right)+k$
• D. None of these

Answer 1

Answer: (C)

$I=\int 2\sin x\cdot \cos x\cdot \log \cos xdx$
put $\log \cos x=t$
$\therefore -\frac{\sin x}{\cos x}dx=dt$
$I=\int 2\sin x\cdot \cos x\cdot t\frac{\cos x}{-\sin x}dt$
$=-2\int {\cos }^{2}x.tdt=-2\int t{e}^{2t}dt$
$=-2\left[t\cdot \frac{e^{2t}}{2}-\int \frac{e^{2t}}{2}\cdot dt\right]=-t{e}^{2t}+\frac{1}{2}{e}^{2t}+k$
$=e^{2t}\left(\frac{1}{2}-t\right)+k={\cos }^{2}x\left\{\frac{1}{2}-\log \cos x\right\}+k$