Finding the indefinite integral of sin^2(pi*x) cos^5(pi*x)

[grafs50]

Homework Statement

∫(sin²(πx)*cos⁵(πx))dx.

Homework Equations

Just the above.

The Attempt at a Solution

I have no idea how pi effects the answer, so I basically solved ∫(sin²(x)^cos⁵(x))dx.

∫(sin²(x)*cos⁴(x)*cos(x))dx
∫sin²(x)*(1-sin²(x))²*cos(x))dx
U-substitution
u = sin x du = -cos x dx

∫(u²*(1-u²)²) -du
-∫(u²*(1-u²)²) du
-∫(u²*(1-2u²+u⁴) du
-∫(u²-2u⁴+u⁶) du

-((u³/3)-(2u⁵/5)+(u⁷/7).

When I look up what the antiderivative should be on integral-calculator.com. the minus sign outside of the parentheses disappears, which I can't figure out how.

Can anyone give me some help?

 

[RUber]

If u = sin x then du = cos x, not - cos x.

 

[LCKurtz]

Homework Statement

∫(sin²(πx)*cos⁵(πx))dx.

Homework Equations

Just the above.

The Attempt at a Solution

I have no idea how pi effects the answer, so I basically solved ∫(sin²(x)^cos⁵(x))dx.

∫(sin²(x)*cos⁴(x)*cos(x))dx
∫sin²(x)*(1-sin²(x))²*cos(x))dx
U-substitution
u = sin x du = -cos x dx

Fix the sign error that RUber pointed out. Then do the original problem the same way but let $u =\sin(\pi x)$.
[Edit:] Fixed typo.

 

[Mark44]

If u = sin x then du = cos x

No doubt just an oversight, but du = cos(x) dx.

 

[grafs50]

Thanks everyone. Figured it out.