Mastering Integration to Solving 48sin^4(3x) with Ease

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I think I've done it at least once every day for the past week.
I week getting it wrong on the homework site.
Worst off all, it only alows 5 entries. I've used 4. Please help.
 
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Just curious, are you taking the anti-derivative, or a definite integral?

If it's a definite integral, it looks like a textbook case for Wallis' formula.
 
Use this formula \int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx
 
noblegas's formula is probably simplest but the general method for even powers of sine or cosine is to use the double angle formula: cos(2x)= 1- 2sin2(x)= 2cos^2(x)- 1. From that sin^2(x)= (1/2)(1- cos(2x)) and cos^2(x)= (1/2)(1+ cos(2x)).

For a fourth power, use that twice: sin^4(x)= ((1/2)(1- cos(2x)))^2= (1/4)(1- cos(2x)- cos^2(2x)). 1/4 and (1/4) cos(2x) can be integrated directly. To integrate (1/4)cos2(2x) do it again: (1/4)cos^2(2x)= (1/4)(1+ cos(4x)).
 
Also, if you want to check your answer before entering it on the homework site, you can always post your concerns here.
We're not going to give you the answer, but if you show what you did we can always take a look.
 
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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