Solve this: Q I = ∫0πx sin2x sinπ2cos x2x-πdx - Maths - Integrals

Question

Solve this:
Q I = ∫ 0 π x   sin 2 x   sin π 2 cos
  x 2 x - π d x

Lovina Kansal · Accepted Answer

let I=∫0πxsin2x sinπ2cosx2x-πdxLet 2x-π=y⇒2dx=dy⇒dx=12dyAt x=0, y=-πx=π,y=πand use x=y+π2=π2+y2So,I=14∫-πππ+y2sinπ+ysinπ2cosπ2+y2ydy=14∫-πππ+y-sinysin-π2siny2ydy=14∫-πππsinyy×sinπ2siny2dy+14∫-ππysinyy×sinπ2siny2dyLet f(y)=πsinyy×sinπ2siny2f(-y)=πsin(-y)(-y)×sinπ2sin-y2=-πsinyysinπ2siny2=-f(y)∴f(y)=-f(-y)⇒f(y) is an odd function⇒∫-ππf(y) dy=0∴I=14∫-ππsiny sinπ2siny2 dyLet g(y)=siny sinπ2siny2⇒g(-y)=sin-y sinπ2sin-y2⇒g(-y)=siny sinπ2siny2⇒g(-y)=g(y)⇒g(y) is an even function⇒∫-ππ g(y) dy=2∫0πg(y)dySO,I=24∫0πsiny sinπ2siny2dy=12∫0π2siny2cosy2sinπ2siny2dyLet siny2=t⇒12cosy2dy=2dtWhen y=0,t=sin0=0y=π,t=sinπ2=1So,I=∫01tsinπt2×2dt=2-t cosπt2×2π01-∫01-cosπt2×2πdt=2-cosπ2×π2+4π∫01cosπt2dt=4π∫01cosπt2dt=4π×2π[sinπt2]01=8π2sinπ2-sin0=8π2

Solve this: Q. I = ∫ 0 π x sin 2 x sin π 2 cos x 2 x - π d x

Solve this:
Q. I = $\int_{0}^{π} \frac{x \sin 2 x \sin (\frac{π}{2} \cos x)}{2 x - π} d x$