Question

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

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\pi$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

Solution:

Let $I=A=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x+\cos x}............\left(i\right)$

Let $I=B=\stackrel{\pi }{\underset{0}{\int }}\frac{\sin xdx}{\sin x-\cos x}............\left(ii\right)$

$[\because \stackrel{a}{\underset{0}{\int }}f(x)dx=\stackrel{a}{\underset{0}{\int }}f(a-x\left)dx\right]$

On adding eqn (i) and (ii), we get

$2I=\stackrel{\pi }{\underset{0}{\int }}(\frac{\sin x}{\sin x+\cos x}+\frac{\sin x}{\sin x-\cos x})dx$

or, $2I=\stackrel{\pi }{\underset{0}{\int }}\frac{{\sin }^{2}x-\sin x\cos x+{\sin }^{2}x+\sin x\cos x}{{\sin }^{2}x-{\cos }^{2}x}dx$

or, $2I=4\stackrel{\frac{\pi }{2}}{\underset{0}{\int }}\frac{{\sin }^{2}x}{{\sin }^{2}x-{\cos }^{2}x}dx..............\left(iii\right)$

$[\because \stackrel{2a}{\underset{0}{\int }}f(x)dx=2\stackrel{a}{\underset{0}{\int }}f(x\left)dx\right]$

or, $2I=4\stackrel{\frac{\pi }{2}}{\underset{0}{\int }}\frac{{\cos }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}dx..............\left(iv\right)$

$[\because \stackrel{a}{\underset{0}{\int }}f(x)dx=\stackrel{a}{\underset{0}{\int }}f(a-x\left)dx\right]$

On adding eqn.(iii) and (iv), we get,

or, $4I=4\stackrel{\frac{\pi }{2}}{\underset{0}{\int }}\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x-{\cos }^{2}x}dx$

or, $4I=4[x{]}_0^{\frac{\pi }{2}}$

or, $4I=4\times \frac{\pi }{2}$

or, $I=\frac{\pi }{2}$

So, $B=\frac{\pi }{2}$

Hence, B is the correct option.