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Question

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


Solution

π20sin32sin32x+cos32xdx.I=π20sin32(π2x)sin32(π2x)+cos32(π2x)dx    (a0f(x)dx=a0f(ax)dx)=π20cos32xcos32x+sin32X[sin(π1X)=cos x and cos(pi2x)=sinx]
 on adding Eqs. (i) and (ii), we get
2I=π20sin32+cos32sin32+cos32xdx=π201dx=[x]π20=π20I=π4

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