Trig Integral

**VitaX** · March 19th 2010, 07:50 PM

$\int sin^2(2x)cos^3(2x) dx$

$\int sin^2(2x)cos^2(2x)cos(2x) dx$

$\int sin^2(2x)(1-sin^2(2x))cos(2x) dx$

let $u = sin(2x)$ $du = 2cos(2x)dx \rightarrow \frac{1}{2}du = cos(2x)dx$

$\frac{1}{2}\int u^2(1-u^2)du = \frac{1}{2} \int u^2 - u^4 du$

$\frac{1}{2}\left[\frac{u^3}{3} - \frac{u^5}{5}\right] + C \rightarrow \frac{1}{2}\left[\frac{5u^3 - 3u^5}{15}\right] + C$

$=\frac{1}{6}sin^3(2x) - \frac{1}{10}sin^5(2x) + C$

I thought I did everything right but wolfram got a different answer than me. Is what I did here correct?

**Soroban** · March 19th 2010, 08:00 PM

Hello, VitaX!

Your work is correct.

What was wolfram's answer?

**VitaX** · March 19th 2010, 08:06 PM

Originally Posted by Soroban

Hello, VitaX!

Your work is correct.

What was wolfram's answer?

$\frac{1}{16}sin(2x) - \frac{1}{96}sin(6x) - \frac{1}{160}sin(10x) + C$

**drumist** · March 19th 2010, 09:02 PM

Wolfram is telling me that both solutions are equal. You can probably get from one solution to the other using a lot of double angle formulas and the like.

In the meantime, rest assured that your solution is correct.

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