Definite Integral (Calculus) this is a revision problem, not a homework problem. Sincere thanks for help.
Question: Evaluate $\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}$.
The answer is $\pi/4$, but I am unable to work out the method.
Sincere thanks for help.
 A: Notice that
$$\frac{d}{dx} (\log(\cos x + \sin x)) = \frac{\cos x - \sin x}{\cos x + \sin x}.$$
Now if
$$I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx= \int_0^{\pi/2} 1 - \frac{\sin x}{\cos x + \sin x}dx$$
then we see that
$$2I = \int_0^{\pi/2} 1 + \frac{\cos x - \sin x}{\cos x + \sin x} dx = \pi/2+0.$$
A: Let $f(x)=\dfrac{\cos x}{\cos x+\sin x}$ and $g(x)=\dfrac{\sin x}{\cos x+\sin x}$.  Note that $g\left(\frac{\pi}{2}-x\right)=f(x)$, which implies that $\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}g(x)dx$.  Also, $f(x)+g(x)=1$ for all $x$, so $\int_0^{\pi/2}(f(x)+g(x))dx=\pi/2$.  
Consequently,
$\frac{\pi}{2}=\int_0^{\pi/2}(f(x)+g(x))dx=\int_0^{\pi/2}f(x)dx+\int_0^{\pi/2}g(x)dx=2\int_0^{\pi/2}f(x)dx.$
A: $$I = \int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}dx \tag{1}$$
$$I = \int_0^{\pi/2}\frac{\cos( \frac{\pi}{2}-x)}{\cos( \frac{\pi}{2}-x)+sin( \frac{\pi}{2}-x)}dx = \int_0^{\pi/2}\frac{\sin x}{\cos x+\sin x}dx \tag{2}$$
Add (1) and (2) to get  
$$2I = \int_0^{\pi/2}\frac{\cos x + \sin x}{\cos x+\sin x}dx =\int_0^{\pi/2}dx = \pi/2$$
Thus, $I$  = $\pi/4$
A: Another easy way would be to directly write that:
$$\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}=\int_0^{\pi/2}\frac{1}{2}\left(1+\frac{\cos x - \sin x}{\cos x+\sin x}\right)=\frac{1}{2}(x+\ln(\sin x +\cos x))\big|_0^{\pi/2} = \frac{\pi}{4}.$$
Q.E.D.
A: Let $ I:= \displaystyle{ \int_{0}^{\pi /2} \frac{\cos x}{\cos x+ \sin x } dx}$ and $ J:= \displaystyle{ \int_{0}^{\pi /2} \frac{\sin x}{\cos x+ \sin x } dx}$.
Now we can see that: $ I+J =\frac{\pi}{2}$ and $ \displaystyle{I-J= \int_{0}^{\pi /2} \frac{ \cos x - \sin x}{\cos x + \sin x} dx = \log ( \sin x +\cos x) |_0^{\pi /2} } =0$
Solving this algebraic system we get that $ \boxed{ I=J= \frac{\pi}{4}}$.
