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Evaluate the integral:

$$\int \sin^4 2x \cos 2x\, dx$$

What I currently did is

$$\int \left(\frac{1-\cos 2x}{2}\right)^2 \cos2x\, dx$$

Honestly I'm not sure what to do, I was absent from class that day. If I try substitution I get $du$ with $\sin(2x)$ and you cant mix variables.. Any help would be appreciated.

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  • $\begingroup$ Try and derive $\sin^5(2x)$. $\endgroup$ – Yves Daoust Oct 5 '15 at 19:04
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Notice, Let $$\sin 2x=t\implies 2\cos 2x dx=dt\implies \cos 2x \ dx=\frac{dt}{2}$$ $$\int \sin^4 2x\cos 2x\ dx=\int t^4\frac{dt}{2}$$ $$=\frac{1}{2}\int t^4dt=\frac{t^5}{10}+C$$ substituting $t=\sin 2x$ $$=\color{blue}{\frac{\sin^5 2x}{10}+C}$$

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    $\begingroup$ Hem, $t^5/10$ maybe... $\endgroup$ – Yves Daoust Oct 5 '15 at 19:06
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    $\begingroup$ Thank you all, I was thinking waaay too hard. I was trying to manipulate the original equation when I could have used substitution right off the bat. Thank you again $\endgroup$ – bankey Oct 5 '15 at 19:09
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$$\frac{1}{2}\int \left(\sin^4 2x\right)\left(\cos 2x\right)d(2x)=\frac{1}{2}\int \left(\sin^4 2x\right)d(\sin 2x)$$

Now let $u=\sin 2x$. The integral becomes

$$\frac{1}{2}\int u^4\ du=\frac{1}{2}\left(\frac{u^5}{5}+C_1\right)=\frac{(\sin 2x)^5}{10}+C_2$$

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  • $\begingroup$ Not getting the hint, the way its explained in the book is different than what you are posting $\endgroup$ – bankey Oct 5 '15 at 19:05
  • $\begingroup$ @bankey Please tell me which step you're not getting. All the integrals here are equal to the original integral. $\endgroup$ – user236182 Oct 5 '15 at 19:47
  • $\begingroup$ I was not understanding the hint, but after I went through Harish's answer, and when you elaborated your answer, I completely understand now, thank you so much :) $\endgroup$ – bankey Oct 5 '15 at 19:57

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