Question

Consider the integrals
$A=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x+\cos x}$ and $B=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x-\cos x}$
Which one of the following is correct?

• A. $A=2B$
• B. $B=2A$
• C. $A=B$
• D. $A=3B$

Answer 1

The correct option is C $A=B$

Solution:

Given that:

$A=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x+\cos x}$ and $B=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x-\cos x}$

Now, $A=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin (\pi -x)dx}{\sin (\pi -x)+\cos (\pi -x)}$ .... $[\because \stackrel{a}{\underset{0}{\int }}f(x)dx=\stackrel{a}{\underset{0}{\int }}f(a-x\left)dx\right]$

$=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x-\cos x}=B$

$\therefore A=B$

Hence, C is the correct option.