Simplifying Trigonometric Integrals

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Homework Statement



Evaluate the integral.

Homework Equations



\int sin^2(\pi x) cos^5 (\pi x) dx

The Attempt at a Solution



I tried first by splitting the cosine up

\int sin^2(x) [1-cos^2(x)] cos^2(x) cos(x) dx and from there use u-substitution. However, I am not sure what to substitute. Any ideas?
 
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Yae Miteo said:

Homework Statement



Evaluate the integral.

Homework Equations



\int sin^2(\pi x) cos^5 (\pi x) dx

The Attempt at a Solution



I tried first by splitting the cosine up

\int sin^2(x) [1-cos^2(x)] cos^2(x) cos(x) dx and from there use u-substitution. However, I am not sure what to substitute. Any ideas?


$$ \int sin^2(\pi x) cos^5 (\pi x) dx $$
$$ = \int sin^2(\pi x) cos^4 (\pi x) cos(\pi x) dx $$
$$ = \int sin^2(\pi x) (1 - sin^2(\pi x))^2 cos(\pi x) dx $$

Since the power of sin was even and cos was odd, you should save a factor of ##cos(x)## and convert the remaining ##cos^2(x)## terms to their ##1 - sin^2(x)## equivalents.

Can you see a substitution from here that would help?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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