Question

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

Answer 1

$$LetI={\int }_0^{\frac{\pi }{2}}(a^2{\cos }^{2}x+b^2{\sin }^{2}x)dx.Then,$$
$$I={\int }_0^{\frac{\pi }{2}}(a^2{\cos }^{2}x+b^2(1-{\cos }^{2}x))dx$$
$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}(b^2+(a^2-b^2){\cos }^{2}x)dx$$
$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}(b^2+\frac{(a^2-b^2)(1+\cos 2x)}{2})dx$$
$$\Rightarrow I={[b^{2}x+\frac{a^2-b^2}{2}(x+\frac{\sin 2x}{2})]}_0^{\frac{\pi }{2}}$$
$$\Rightarrow I=\frac{b^2\pi }{2}+\frac{a^2-b^2}{2}\frac{\pi }{2}+0$$
$$\Rightarrow I=\frac{\pi }{4}(a^2+b^2)$$