Tough integral

**griffsterb** · June 24th 2008, 05:52 PM

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.

**Jhevon** · June 24th 2008, 05:59 PM

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 $\sin t \tan t = \sin t \cdot \frac {\sin t}{\cos t} = \frac {\sin^2 t}{\cos t} = \frac {1 - \cos^2 t}{\cos t}= \sec t - \cos t$ . you should know the integrals of each of those

**griffsterb** · June 24th 2008, 06:07 PM

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

**Jhevon** · June 24th 2008, 06:09 PM

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!

**polymerase** · June 24th 2008, 06:14 PM

Or you can apply by parts:

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

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

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

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

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