Question

Evaluate: ${\int }_0^{\pi /2}\frac{8\sin \theta +4\cos \theta }{\sin \theta +\cos \theta }$.

Answer 1

$I={\int }_0^{\pi /2}\frac{8\sin \theta +4\cos \theta }{\sin \theta +\cos \theta }d\theta$

$I={\int }_0^{\pi /2}\frac{8\sin (\pi /2-\theta )+4\cos (\pi /2-\theta )}{\sin (\pi /2-\theta )+\cos (\pi /2-\theta )}d\theta$

$={\int }_0^{\pi /2}\frac{8\cos \theta +4\sin \theta }{\sin \theta +\cos \theta }d\theta$

$2I={\int }_0^{\pi /2}12\left(\frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }\right)d\theta$

$=12{]}_0^{\pi /2}$

$=12.\pi /2$

$\Rightarrow I=\frac{6\pi }{2}=3\pi$