Using the substitution x → a + b − x

Idea. The integral looks hard because no antiderivative is obvious, but the integrand is paired with another one that has the same integral.

Computation. The substitution $u=\frac\pi2-x$ turns $\sin x$ into $\cos u$ and vice versa, so
$$I=\int_0^{\pi/2}\frac{\cos u}{\cos u+\sin u}\,du=:J.$$
Adding,
$$I+J=\int_0^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}\,dx=\frac\pi2.$$
Since $I=J$, we get $I=\dfrac\pi4$.

General $p$. The same substitution exchanges $\sin^p$ and $\cos^p$, so the integral $I_p$ equals $\int_0^{\pi/2}\frac{\cos^px}{\sin^px+\cos^px}dx$ and $2I_p=\frac\pi2$.

Remark. The general principle is $\int_a^bf(x)\,dx=\int_a^bf(a+b-x)\,dx$. It is a recurrent trick in olympiad-style integrals.