Question

Integrate:

${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}dx$ Then

• A. $I=\frac{\pi }{3}$
• B. $I=\frac{\pi }{4}$
• C. $I=\frac{\pi }{6}$
• D. $I=\frac{\pi }{5}$

Answer 1

Answer: (B)

$I=\frac{\pi }{4}$

We have,

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}dx$ …….. (1)

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\cos x+\sin x}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\sin x+\cos x}dx$ …….. (2)

On adding equation (1) and (2), we get

$2I={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\sin x+\cos x}dx+{\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}dx$

$2I={\int }_0^{\frac{\pi }{2}}\frac{sinx+\cos x}{\sin x+\cos x}dx$

$2I={\int }_0^{\frac{\pi }{2}}\left(1\right)dx$

$2I={\left(x\right)}_0^{\frac{\pi }{2}}$

$2I=(\frac{\pi }{2}-0)$

$I=\frac{\pi }{4}$

Hence, this is the answer.