Evaluate $$ \int_0^\frac{\pi}{2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta.\qquad\text{(1)} $$
By letting $t=\tan\theta$, $(1)$ equals $$ \int_0^\infty\frac{1}{(1+t)(1+t^2)} \,dt,$$ and then?
By letting $t=\tan\frac{\theta}{2}$, (1) equals $$ \int_0^1\frac{(1-t^2)\cdot2}{(1+t^2)\cdot(1-t^2+2t)}\,dt, $$ and then?
All need to much effort.
Use symmetry instead. Notice under the change of variable $\theta \mapsto \frac{\pi}{2}-\theta$
$$I = \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\cos\theta + \sin\theta}d\theta$$
by the trig identity $\cos\left(\frac{\pi}{2}-\theta\right) = \sin\theta$ and vice-versa. Then add the two integrals:
$$2I = \int_0^{\frac{\pi}{2}} \frac{\cos\theta}{\cos\theta + \sin\theta}d\theta + \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\cos\theta + \sin\theta}d\theta = \int_0^{\frac{\pi}{2}}d\theta = \frac{\pi}{2}$$
Therefore $I = \frac{\pi}{4}$
