Question

$I_1=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x-\cos x}{1+\sin x\cos x}dx,I_2=\underset{0}{\overset{2\pi }{\int }}{\cos }^{6}xdx$,
$I_3=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{3}xdx,I_4=\underset{0}{\overset{1}{\int }}\ln (\frac{1}{x}-1)dx$, then

• A. $I_2=I_3=I_4=0,I_1\ne 0$
• B. $I_1=I_2=I_3=0,I_4\ne 0$
• C. $I_1=I_3=I_4=0,I_2\ne 0$
• D. $I_1=I_2=I_4=0,I_3\ne 0$

Answer 1

The correct option is C $I_1=I_3=I_4=0,I_2\ne 0$

$I_1=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x-\sin x}{1+\sin x\cos x}dx=-I_1$
$\Rightarrow I_1=0$
$I_3=0$ as ${\sin }^{3}x$ is odd
$I_4=\underset{0}{\overset{1}{\int }}$$\ln \left(\frac{1-x}{x}\right)dx$
$=\underset{0}{\overset{1}{\int }}$$\ln \left(\frac{1-(1-x)}{1-x}\right)dx$
$=\underset{0}{\overset{1}{\int }}$$\ln \frac{x}{1-x}dx=-I_4$
$\Rightarrow I_4=0$
$I_2=\underset{0}{\overset{2\pi }{\int }}{\cos }^{6}xdx=2{\int }_0^{\pi }{\cos }^{6}xdx\ne 0\{\because \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{when}f\left(x\right)=f(2a-x)\}$