Question

# By using properties of definite integrals, evaluate the integrals ∫π20sin32sin32x+cos32xdx.

Solution

## ∫π20sin32sin32x+cos32xdx.⇒I=∫π20sin32(π2−x)sin32(π2−x)+cos32(π2−x)dx    (∵∫a0f(x)dx=∫a0f(a−x)dx)=∫π20cos32xcos32x+sin32X[∵sin(π1−X)=cos x and cos(pi2−x)=sinx]  on adding Eqs. (i) and (ii), we get 2I=∫π20sin32+cos32sin32+cos32xdx=∫π201dx=[x]π20=π2−0⇒I=π4

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