Simple integral?

Question

I want to calculate the integral of sin^3(x)*cos^5(x)dx (the power are meant to be (sin(x))^n).

I know there are some reduction formulas, but I don't like them. I found a simple way, but it's probably wrong and I don't understand where's the error.

I simply set sin x = t, thus cos x dx = dt.
The sin^3(x) is t^3, one of the 5 cos x becomes the differential, while I'm left with cos^4 = (1-t^2)^2.

The new integral in t is simply:
t^3(1-t^2)^2 dt.

Now this is an easy polynomial and I can integrate it like usual and then set t = sin(x).

It seems so easy and linear, yet I'm quite sure it doesn't work, but why is it so?

? · Answer

Expand (1-t^2)^2= 1-2t^2 +t^4
The integral will be
t^3 - 2 t^5 +t^7. Integrate it. You will have the answer
1/4 t^4 -1/3 t^6 +1/8 t^8= 1/4 sin^4 x -1/3 sin^5 x
+ 1/8 sin^8 x. This seem to be the answer. You need not to back at the stage that you mentioned.

PiRhoTeXNiX · Answer

I don't understand why the substitution would make it simpler. If you look at Some Body's solution, at the step

∫ (sin³ x - 2sin⁵ x + sin⁷ x) cos x dx

you can practically just integrate in one step from there since the cos(x) term is just the chain-rule multiplier derivative you'd get from differentiating a power of sin(x).

That being said, it turns out that you can get less terms in the final answer by turning the sin(x) powers into cos(x) powers instead of the other way round:

∫ sin³ x cos⁵ x dx
= ∫ sin x (sin² x) cos⁵ x dx 
= ∫ sin x (1 - cos² x) cos⁵ x dx 
= ∫ sin x (cos⁵ x - cos⁷ x) dx
= ⅛ cos⁸ x - ⅙ cos⁶ x + C

Some Body · Answer

That absolutely works.
∫ sin³ x cos⁵ x dx
∫ sin³ x cos⁴ x cos x dx
∫ sin³ x (cos² x)² cos x dx
∫ sin³ x (1 - sin² x)² cos x dx
∫ sin³ x (1 - 2sin² x + sin⁴ x) cos x dx
∫ (sin³ x - 2sin⁵ x + sin⁷ x) cos x dx

If u = sin x, du = cos x dx:
∫ (u³ - 2u⁵ + u⁷) du
¼ u⁴ - ⅓ u⁶ + ⅛ u⁸ + C

¼ sin⁴ x - ⅓ sin⁶ x + ⅛ sin⁸ x + C

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Simple integral?

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