∫sin^3 x . cos^4 x dx How to do??

Question

it is sin to the power of 3, not sin to the power of 3x. and cos to the power of 4, not cos to the power of 4x.

let u=cos x

Answer 1

∫ (sin^2 x)(sin x)(cos^4 x) dx
∫ (1 - cos^2 x)(sin x)(cos^4 x) dx

Let, u = cosx,
du = sinx dx

∫ (1 - u^2)(u^4) du
∫ (u^4 - u^6) du
= 1/5 (u^5) - 1/7 (u^7) + c
= 1/5 cos^5(x) - 1/7 cos^7(x) + c

Answer 2

∫sin^3 x . cos^4 x dx
= ∫sin x (sin^2 x) (cos^4 x) dx
= ∫sin x (1-cos^2 x) (cos^4 x) dx

Then, let u=cos x, du=-sin x dx.

The integral then becomes
∫sin^3 x . cos^4 x dx = ∫-(1-u^2)u^4 du = ∫(u^6-u^4) du

Evidently, it is managable from here. Just don't forget to return the equation back in terms of x.

Answer 3

I = ∫ (sin ² x) (cos^(4) x) (sin x) dx
I = ∫ (1 - cos ² x) (cos^(4) x) (sin x) dx
let u = cos x
du = - sin x dx
I = - ∫ (1 - u ²) (u^4) du
I = ∫ u^6 - u^4 du
I = u^7 / 7 - u^5 / 5 + C
I = cos^(7) x / 7 - cos^(5) x / 5 + C