The definite integral of the square of SiNx =? How do you calculate 0-90,45? I can't

The definite integral of the square of SiNx =? How do you calculate 0-90,45? I can't

The above formula is equal to 0.5 (1-cos2x) and 0.5x + sin2x + C is obtained after integration
Definite integral should have upper and lower limits

For the definite integral of the square of function x (SiNx), the lower limit is 0 and the upper limit is 1

A:
Because ∫ xsin & # 178; X DX
=∫x(1-cos2x)/2 dx
=1/2∫x(1-cos2x) dx
=1/2∫x-xcos2x dx
=1/2(∫x dx - ∫xcos2x dx)
=x²/4-1/4xsin2x+1/4∫sin2x dx
=x²/4-1/4xsin2x-1/8cos2x + C
So ∫ (0 to 1) xsin & # 178; X DX
=X & # 178 / / 4-1 / 4xsin2x-1 / 8cos2x | (0 to 1)
=1/4-sin2/4-cos2/8-(0-0-1/8)
=3/8-sin2/4-cos2/8

Finding the square of SiNx times the lower limit 0 and upper limit π of X definite integral

X (SiNx) ^ 2 = x * (1-cos2x) / 2 = 1 / 2 * X-1 / 2xcos2x ∫ x (SiNx) ^ 2DX = 1 / 2 ∫ xdx - 1 / 2 ∫ xcos2xdx = 1 / 4x ^ 2 - 1 / 4 ∫ xdsin2x = 1 / 4x ^ 2-1 / 4 (xsin2x - ∫ sin2xdx) = 1 / 4x ^ 2-1 / 4xsin2x + 1 / 4 ∫ sin2xdx = 1 / 4xsin2x ^ 2-1 / 4xsin2x-1 / 8cos2x + C to carry π in

Finding the definite integral of X / 1 + SiNx from 0 to π

Let x = π - u, DX = - Dux = 0, u = π x = π, u = 0n = ∫ (0 → π) x / (1 + SiNx) DX = ∫ (π → 0) (π - U) / [1 + sin (π - U)] * (- Du) = ∫ (0 → π) (π - U) / (1 + sinu) Du = ∫ (0 → π) (π - x) / (1 + SiNx) DX = π∫ (0 →