Evaluate the following Integral: ∫(a^2cos^2x+ b^2sin^2x)dx, x ∈ [0, π/4]

Question

Evaluate the following Integral:

\(\int\limits_0^{\pi/4} \)(a²cos²x + b²sin²x) dx

Answer 1

I =  \(\int\limits_0^{\pi/4} \)(a²cos²(x) + b²sin²(x)) dx

We know integral of cosx is sinx

sin 0 = 0, sin 90 = 1

= (b² + a²)\(\cfrac{\pi}8+\cfrac{(b^2-a^2)}4\)