1. ## Tough integral

Well, probably not so tough. I haven't done much integrating in a while and I ran into an integral of (sint*tant) in my homework. I couldn't rearrange it in any helpful way, so I'm hoping there is an identity for it. Thanks.

2. Originally Posted by griffsterb
Well, probably not so tough. I haven't done much integrating in a while and I ran into an integral of (sint*tant) in my homework. I couldn't rearrange it in any helpful way, so I'm hoping there is an identity for it. Thanks.
note that $\displaystyle \sin t \tan t = \sin t \cdot \frac {\sin t}{\cos t} = \frac {\sin^2 t}{\cos t} = \frac {1 - \cos^2 t}{\cos t}= \sec t - \cos t$. you should know the integrals of each of those

3. Bah, should have seen that sin squared identity in the numerator! Shakin the rust off :) Thanks so much!

4. Originally Posted by griffsterb
Bah, should have seen that sin squared identity in the numerator! Shakin the rust off Thanks so much!
you're welcome

keep at it!

5. Or you can apply by parts:

$\displaystyle u=\tan x \to du=\sec^2x$ $\displaystyle dv=\sin x \to v=-\cos x$

$\displaystyle \int \sin x\tan x=-\cos x\tan x+\int \cos x\sec^2 x\;dx$
$\displaystyle \sec^2x\cos x = \frac{\cos x}{\cos^2x}=\sec x$

Therefore, $\displaystyle \int \cos x\sec^2 x\;dx=\int \sec x \;dx=\ln|\sec x+\tan x|+C$

Thus, $\displaystyle \int \sin x\tan x\;dx=\ln|\sec x+ \tan x|-\sin x +C$