Solution to Problem 0201 | Integration

Problem 201

Evaluate $\int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx}$

Solution:

let $I = \int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx}$

=$\int\limits_0^{\pi /2} {{{\cos (\pi /2 - x)} \over {\sin (\pi /2 - x) + \cos (\pi /2 - x)}}dx}$

=$\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx}$

So $\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx}  = I$

Tharefore

2I=I+I

=$\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx}$ +$\int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx}$

=$\int\limits_0^{\pi /2} {{{\sin x + \cos x} \over {\cos x + \sin x}}dx}$

=$\int\limits_0^{\pi /2} {dx}$

=${\left[x\right]}\nolimits_0^{\pi /2}={\pi \over 2}$

$\therefore$ I=${\pi \over 4}$