1. ## Integration Question

I feel like I should know how to do this but I'm kind of getting stuck

Question: $\displaystyle \int^{\pi/4}_0 \frac{e^{tanx}}{1-sin^2 x}$

So i used the trig identity and got
$\displaystyle \int^{\pi/4}_0 \frac{e^{tanx}}{cos^2 x}$

$\displaystyle let \:u=tanx \:\:\: du=\frac{1}{cosx}$

$\displaystyle \int^{\pi/4}_0 \frac{e^u}{cosx\, du}$

so basically I am left with a spare 1/cosx that I'm not sure what to do with, any help would be appreciated

2. Originally Posted by jfinkbei

so basically I am left with a spare 1/cosx that I'm not sure what to do with, any help would be appreciated
No you're not

$\displaystyle u=\tan x \implies du=\frac{1}{\cos^2x}$

3. Haha, I knew that... thanks a lot though! could have saved myself some time writing latex had I payed more attention...

4. Originally Posted by pickslides
No you're not

$\displaystyle u=\tan x \implies du=\frac{{\color{red}dx}}{\cos^2x}$
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5. Originally Posted by Miss
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It's usually implied that it's wrt $\displaystyle x$ if $\displaystyle dx$ is not specified.

For the OP you can make life easier for yourself by changing your limits to be in terms of u.

$\displaystyle u = \tan \left(\frac{\pi}{4}\right) = 1$

$\displaystyle u = \tan (0) = 0$

So the integral is now $\displaystyle \int ^1_0 \, e^u \, du$