Question

\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) d x . Then, \]
\[ I = \int_0^\frac{\pi}{2} \left( a^2 \cos^2 x + b^2 \left( 1 - \cos^2 x \right) \right) d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( b^2 + \left( a^2 - b^2 \right) \cos^2 x \right) dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( b^2 + \frac{\left( a^2 - b^2 \right)\left( 1 + \cos 2x \right)}{2} \right)dx\]
\[ \Rightarrow I = \left[ b^2 x + \frac{a^2 - b^2}{2}\left( x + \frac{\sin 2x}{2} \right) \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{b^2 \pi}{2} + \frac{a^2 - b^2}{2}\frac{\pi}{2} + 0\]
\[ \Rightarrow I = \frac{\pi}{4}\left( a^2 + b^2 \right)\]