Question

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

Answer 1

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]
\[ = \int_0^\frac{\pi}{4} \left[ a^2 \left( \frac{1 + \cos2x}{2} \right) + b^2 \left( \frac{1 - \cos2x}{2} \right) \right]dx\]
\[ = \int_0^\frac{\pi}{4} \left[ \left( \frac{a^2 + b^2}{2} \right) + \left( \frac{a^2 - b^2}{2} \right)\cos2x \right]dx\]
\[ = \left( \frac{a^2 + b^2}{2} \right) \int_0^\frac{\pi}{4} dx + \left( \frac{a^2 - b^2}{2} \right) \int_0^\frac{\pi}{4} \cos2xdx\]

\[= \left.\left( \frac{a^2 + b^2}{2} \right) \times x\right|_0^\frac{\pi}{4} + \left.\left( \frac{a^2 - b^2}{2} \right) \times \frac{\sin2x}{2}\right|_0^\frac{\pi}{4} \]
\[ = \left( \frac{a^2 + b^2}{2} \right)\left( \frac{\pi}{4} - 0 \right) + \left( \frac{a^2 - b^2}{4} \right)\left( \sin\frac{\pi}{2} - \sin0 \right)\]
\[ = \left( \frac{a^2 + b^2}{2} \right)\frac{\pi}{4} + \left( \frac{a^2 - b^2}{4} \right)\left( 1 - 0 \right)\]
\[ = \left( a^2 + b^2 \right)\frac{\pi}{8} + \frac{1}{4}\left( a^2 - b^2 \right)\]