Test at 7pm, stuck on Trig Integrals.

**UDaytonFlyer** · March 4th 2008, 11:46 AM

There are two problems on our practice test/solution that I just can't seem to grasp. The first is: Int(tan^2 (x)dx) ... The solution is tanx-x+c, but I don't understand how the prof. got there.

Second: Int(sin^3 (x) cos^2 (x)dx) I know that I need to "split up" the sin^3 into a sin^2 and a sin to get a (1-cos^2 (x)) term, but after that I'm stuck...

Thanks for any help you can provide. I'm just having a hard time grasping the concept/steps of trig integrals in general. Argh.

Thanks again!

**Soroban** · March 4th 2008, 12:33 PM

Hello, UDaytonFlyer

$\int\tan^2\!x\,dx$

Identity: . $\sec^2\!\theta \:=\:\tan^2\!\theta +1$

We have:. $\int\tan^2\!x\,dx \;=\;\int(\sec^2\!x-1)\,dx \;=\;\tan x - x + C$

$\int \sin^3\!x\cos^2\!x\,dx$

We have: . $\int\sin^2\!x\cos^2\!x(\sin x\,dx) \;=\;\int(1-\cos^2\!x)\cos^2\!x(\sin x\,dx)$

. .. $=\;\int(\cos x - \cos^3\!x)(\sin x\,dx)$

Then let $u \:=\:\cos x\quad\Rightarrow\quad du\:=\:-\sin x\,dx$

. . and we have: . $-\int(u-u^3)\,du \quad \hdots\quad \text{Got it?}$

**UDaytonFlyer** · March 4th 2008, 12:36 PM

I think I get it now... SO the purpose of the trig integral is basically to make it into something easier to integrate, hence the saving of one "factor" of the trig function to use in the dx... Thanks for the help!

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