Answer
The solution is $$\int_{0}^{\pi/2}\frac{1}{1+\sin\theta+\cos\theta}d\theta=\ln2.$$
Work Step by Step
To solve the integral:
$$\int_{0}^{\pi/2}\frac{1}{1+\sin\theta+\cos\theta}d\theta$$
we will use substitution $\tan\frac{\theta}{2}=t\Rightarrow d\theta=\frac{2}{1+t^2}dt,\cos\theta=\frac{1-t^2}{1+t^2}, \sin\theta=\frac{2t}{1+t^2}$
and the integration bounds become $t=0$ and $t=1$. Now putting all this into the integral we have:
$$\int_{0}^{\pi/2}\frac{1}{1+\sin\theta+\cos\theta}d\theta=
\int_{0}^{1}\frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=
\int_{0}^{1}\frac{1}{\frac{1+t^2+2t+1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=
\int_{0}^{1}\frac{1+t^2}{2+2t}\frac{2}{1+t^2}dt=
\int_{0}^{1}\frac{1}{1+t^2}dt=\left.\ln|1+t|\right|_{0}^{1}=\ln2-\ln1=\ln2-0=\ln2.$$
