integrate please quick?

I = ∫ sin ² x ( cos x )^5 dx

let u = sin x

du = cos x dx

I = ∫ ( sin ² x) ( cos ² x ) ( cos ² x ) ( cos x ) dx

I = ∫ ( u ² ) ( 1 - u ² ) ( 1 - u ² ) du

I = ∫ ( u ² ) ( 1 - 2 u ² + u^4) du

I = ∫ ( u^2 - 2 u^4 + u^6) du

I = u^3 / 3 - 2u^5 / 5 + u^7 / 7 + C

I = sin^3 x / 3 - 2 sin^5x / 5 + sin^7x / 7 + C

∫sin^2 x cos^5 x dx

=∫sin² x (1−sin²x)² cos x dx

=∫sin² x (1+sin^4x−2sin²x) cos x dx

=∫sin² x cos x dx+∫sin^6x cos x dx−2∫sin^4 x cos x dx

=sin^3x/3 +sin^7x/7 −2sin^5 x/5 +C

y '=(1-cos^2x)cos^5x=-cos^7x +cos^5x

y=-1/8 sin^8x +1/6 sin^6 x +C

God bless you.

ʃ sin² x cos⁵ dx = ʃ sin² x cos⁴ x cos x dx

...................... = ʃ sin² x (cos² x)² cos x dx

...................... = ʃ sin² x (1 – sin² x)² cos x dx

...................... = ʃ sin² x (1 – 2sin² x + sin⁴ x) cos x dx

...................... = ʃ (sin² x – 2sin³ x + sin⁵ x) cos x dx

...................... = ʃ sin² x cos x dx – ʃ2sin³ x cos x dx + ʃsin⁵ x cos x dx

...................... = (1/3) sin³ x – (2/4)sin⁴ x + (1/6)sin⁶ x + C

...................... = (1/3) sin³ x – (1/2)sin⁴ x + (1/6)sin⁶ x + C

∫ (sin^2 x cos^5 x) dx = ∫(1 - cos^2x)* cos^5x.dx = ∫(cos^5x - cos^7x)dx

∫cos^5x. dx - ∫cos^7 x.dx .... from here proceed further please.