How should I approach this problem?

**FalconPUNCH!** · Feb 19th 2008, 02:49 PM

We just started learning about Trigonometric Integrals and I was doing pretty good with the odd numbered ones, until I got to the even. I don't really know how to approach this problem.

$\int_{0}^{Pi} sin^{2}t cos^{4}t dt$

Any help would be great.

**FalconPUNCH!** · Feb 19th 2008, 03:31 PM

Well I think I got the answer and it took me about two pages to get it trail and error.

Here's how I approached it:

I substituted $Sin^{2}t$ with $1 - cos^{2}t$

$\int_{0}^{Pi} (1-cos^{2}t) (cos^{2}t) (cos^{2}t)$

I used half angle formula to get

$\frac{1}{8} \int_{0}^{Pi} (1- \frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])$

I expanded everything and got

$\frac{1}{8}\int_{0}^{Pi} 2cos2t + cos^{2}t * cost dt$

I went from there and did a few more steps and got $\frac{29Pi}{128}$

Can anyone check if I did it correct?

**Jhevon** · Feb 19th 2008, 04:18 PM

Originally Posted by FalconPUNCH!

Well I think I got the answer and it took me about two pages to get it trail and error.

Here's how I approached it:

I substituted $Sin^{2}t$ with $1 - cos^{2}t$

$\int_{0}^{Pi} (1-cos^{2}t) (cos^{2}t) (cos^{2}t)$

I used half angle formula to get

$\frac{1}{8} \int_{0}^{Pi} (1- \frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])$

I expanded everything and got

$\frac{1}{8}\int_{0}^{Pi} 2cos2t + cos^{2}t * cost dt$

I went from there and did a few more steps and got $\frac{29Pi}{128}$

Can anyone check if I did it correct?

that is incorrect. the answer should be $\frac {\pi}{16}$

you were pretty much on track about how to start. so you went wrong in the algebra or evaluation somewhere

use the fact that $\sin^2 x = \frac {1 - \cos 2x}2$ and $\cos^2 x = \frac {1 + \cos 2x}2$

so, $\int_0^{\pi} \sin^2 x \cos ^4 x~dx = \frac 18 \int_0^{\pi} (1 - \cos 2x)(1 + \cos 2x)^2~dx = \frac 18 \int_0^{\pi} \sin^2 2x (1 + \cos 2x)~dx$

do a similar manipulation for $\sin^2 2x$ as we did before and continue

**FalconPUNCH!** · Feb 19th 2008, 04:56 PM

Thanks yeah the page got messy so I might have multiplied incorrectly.

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