anonymous
  • anonymous
integrate sin^3(6x)cos^2(6x) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
Hmm so we have an `odd power`, that's really good :)
anonymous
  • anonymous
yeah, you can pull off a sinx
zepdrix
  • zepdrix
\[\Large \int\limits \sin6x \color{royalblue}{\sin^26x}\cos^26x\;dx\]Yah that sounds right peg. We'll want to use our `Square Identity` to convert this blue part to cosines.

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anonymous
  • anonymous
\[\int\limits_{?}^{?} (1-\cos ^{2}6x)\cos ^{2}6xsin6x\]
zepdrix
  • zepdrix
yah looks good. NOW you can do a u-sub ^^ u=cos6x
anonymous
  • anonymous
this is where I'm getting confused \[\int\limits_{}^{} (1-u ^{2}u ^{2} du ?\]
anonymous
  • anonymous
woops, forgot the 2nd parenthesis
zepdrix
  • zepdrix
I think you get a negative on your du right?\[\large u=\cos6x \qquad\to\qquad du=-6\sin6x\;dx \qquad\to\qquad \left(-\frac{1}{6}du\right)=\sin6x\;dx\]
anonymous
  • anonymous
yeah, that's right
zepdrix
  • zepdrix
That wasn't the part you were confused on I guess, I just wanted to make sure :)\[\Large \int\limits(1-u^2)u^2\left(-\frac{1}{6}du\right)\]
zepdrix
  • zepdrix
Next, just distribute the u^2 to each term in the brackets. Then you can apply the power rule as normal.
anonymous
  • anonymous
excellent, just got the answer. I see what I was doing wrong.
zepdrix
  • zepdrix
cool c:
anonymous
  • anonymous
thank you!

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