Question

Evaluate the following integrals:

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\sin x+b\sin x}{\sin x+\cos x}dx$

Answer 1

Let I =${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx$ .....(1)

Then,
$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+b\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right]$
$={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\cos x+b\sin x}{\cos x+\sin x}dx$ .....(2)

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{a\sin x+b\cos x}{\cos x+\sin x}+\frac{a\cos x+b\sin x}{\sin x+\cos x}\right)dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{a\sin x+b\cos x+a\cos x+b\sin x}{\sin x+\cos x}\right)dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(a+b\right)\sin x+\left(a+b\right)\cos x}{\sin x+\cos x}dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(a+b\right)\left(\sin x+\cos x\right)}{\sin x+\cos x}dx \end{gathered}$$
$$\begin{gathered} \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(a+b\right)dx \\ \Rightarrow 2I=\left(a+b\right)\times {\overline{)x}}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow 2I=\left(a+b\right)\times \left(\frac{\mathrm{\pi }}{2}-0\right) \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{2}\left(a+b\right) \\ \Rightarrow I=\frac{\mathrm{\pi }}{4}\left(a+b\right) \end{gathered}$$