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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.
$\displaystyle \int_{0}^{Pi} sin^{2}t cos^{4}t dt $
Any help would be great.
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 $\displaystyle Sin^{2}t$ with $\displaystyle 1 - cos^{2}t$
$\displaystyle \int_{0}^{Pi} (1-cos^{2}t) (cos^{2}t) (cos^{2}t)$
I used half angle formula to get
$\displaystyle \frac{1}{8} \int_{0}^{Pi} (1- \frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])(\frac{1}{2}[1+cos^{2}t])$
I expanded everything and got
$\displaystyle \frac{1}{8}\int_{0}^{Pi} 2cos2t + cos^{2}t * cost dt $
I went from there and did a few more steps and got $\displaystyle \frac{29Pi}{128}$
Can anyone check if I did it correct?
that is incorrect. the answer should be $\displaystyle \frac {\pi}{16}$Quote:
Originally Posted by FalconPUNCH!
you were pretty much on track about how to start. so you went wrong in the algebra or evaluation somewhere
use the fact that $\displaystyle \sin^2 x = \frac {1 - \cos 2x}2$ and $\displaystyle \cos^2 x = \frac {1 + \cos 2x}2$
so, $\displaystyle \int_0^{\pi} \sin^2 x \cos ^4 x~dx = \frac 18 \int_0^{\pi} (1 - \cos 2x)(1 + \cos 2x)^2~dx = \frac 18 \int_0^{\pi} \sin^2 2x (1 + \cos 2x)~dx$
do a similar manipulation for $\displaystyle \sin^2 2x$ as we did before and continue
Thanks yeah the page got messy so I might have multiplied incorrectly.
