Problem 201
Evaluate [tex]\int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx} [/tex]
Solution:
let [tex]I = \int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx} [/tex]
=[tex]\int\limits_0^{\pi /2} {{{\cos (\pi /2 - x)} \over {\sin (\pi /2 - x) + \cos (\pi /2 - x)}}dx} [/tex]
=[tex]\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx} [/tex]
So [tex]\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx} = I[/tex]
Tharefore
2I=I+I
=[tex]\int\limits_0^{\pi /2} {{{\sin x} \over {\cos x + \sin x}}dx} [/tex] +[tex]\int\limits_0^{\pi /2} {{{\cos x} \over {\sin x + \cos x}}dx} [/tex]
=[tex]\int\limits_0^{\pi /2} {{{\sin x + \cos x} \over {\cos x + \sin x}}dx} [/tex]
=[tex]\int\limits_0^{\pi /2} {dx} [/tex]
=[tex]{\left[x\right]}\nolimits_0^{\pi /2}={\pi \over 2}[/tex]
[tex]\therefore[/tex] I=[tex]{\pi \over 4}[/tex]
