Question

If $A={\int }_0^{\pi }\frac{\sin x}{\sin x+\cos x}dx,B={\int }_0^{\pi }\frac{\sin x}{\sin x-\cos x}dx$ then

• A. $A+B=0$
• B. $A=B$
• C. $A=B=\pi /2$
• D. $A=-B=\pi$

Answer 1

Answer: (B), (C)

The correct options are
B $A=B$
C $A=B=\pi /2$

$A={\int }_0^{\pi }\frac{\sin x}{\sin x+\cos x}dx={\int }_0^{\pi }\frac{\sin (\pi -x)}{\sin (\pi -x)+\cos (\pi -x)}dx={\int }_0^{\pi }\frac{\sin x}{\sin x-\cos x}dx=B$
For $A={\int }_0^{\pi }\frac{\sin x}{\sin x+\cos x}dx$
Multiply numerator and denominator by ${\csc }^{3}x$
$A={\int }_0^{\pi }\frac{{\csc }^{2}x}{{\csc }^{2}x+\cot x{\csc }^{2}x}dx={\int }_0^{\pi }\frac{{\csc }^{2}x}{1+\cot x+{\cot }^{2}x+{\cot }^{3}x}dx$
Substitute $t=\cot x\Rightarrow dt=-{\csc }^{2}xdx$
$\therefore A=-2{\int }_{\frac{\pi }{2}}^0\frac{1}{t^3+t^2+t+1}dt$
$=2(-\frac{1}{2}{\int }_{\frac{\pi }{2}}^0\frac{t}{t^2+1}dt-\frac{1}{2}{\int }_{\frac{\pi }{2}}^0\frac{1}{t^2+1}dt-\frac{1}{2}{\int }_{\frac{\pi }{2}}^0\frac{1}{t+1}dt)$
$=2{[\frac{1}{4}\log (t^2+1)-\frac{1}{2}\log (t+1)-\frac{1}{2}{\tan }^{-1}t]}_{\frac{\pi }{2}}^0=\frac{\pi }{2}$