Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\left(a^2{\cos }^{2}x+b^2{\sin }^{2}x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(a^2{\cos }^{2}x+b^2{\sin }^{2}x\right)\mathit{d}x.\mathrm{Then}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(a^2{\cos }^{2}x+b^2\left(1-{\cos }^{2}x\right)\right)\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(b^2+\left(a^2-b^2\right){\cos }^{2}x\right)dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(b^2+\frac{\left(a^2-b^2\right)\left(1+\cos 2x\right)}{2}\right)dx \\ \Rightarrow I={\left[b^{2}x+\frac{a^2-b^2}{2}\left(x+\frac{\sin 2x}{2}\right)\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=\frac{b^2\mathrm{\pi }}{2}+\frac{a^2-b^2}{2}\frac{\mathrm{\pi }}{2}+0 \\ \Rightarrow I=\frac{\mathrm{\pi }}{4}\left(a^2+b^2\right) \end{gathered}$$