Check out some similar questions!

interinfinity · Feb 21, 2008, 06:50 PM

I would like to know how to set up this specific integral...
I don't know how to do an integral symbol on here...

I need the integral of : cos^2(x)tan^5(x)
that's... cosine squared x times tangent fifth x

I have tried doing it by integration by parts, but that won't work because the integral of either part is tricky to get without a graphing calculator.

I'm trying to use trig identities, the only thing I can come up with is breaking tan^5(x) up into secent... OK I admit it I don't have a clue what I'm talking about. I hate trig. If no one responds to this post I won't shed a tear, I hate trig.

galactus · Feb 22, 2008, 04:53 AM

Wow, finally a problem with a little meat to it. Something besides the usual "Help me solve x=4 for x".

A lot of times these things take some insight to see the substitution. Besides, trig is cool.

I was a surveyor for years and used it everyday. I didn't have to integrate trig functions, but I used it.

$\int{cos^{2}(x)tan^{5}(x)}dx$

Make the sub $\frac{sin^{5}(x)}{cos^{5}(x)}=tan^{5}(x)$

Then you get $\int\frac{sin^{5}(x)}{cos^{3}(x)}dx$

$\int\frac{sin(x)sin^{4}(x)}{cos^{3}(x)}dx$

Now, if you let $u=cos(x), \;\ -du=sin(x)dx, \;\ x=arccos(u)$

Note that $sin^{4}(arccos(u))=(u^{2}-1)^{2}$

It whittles down to something fairly easy to integrate.

$-\int\frac{(u^{2}-1)^{2}}{u^{3}}du=-\int{u}du+2\int\frac{1}{u}-\int\frac{1}{u^{3}}du$