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Math Help - Integration Question

  1. #1
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    Integration Question

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

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

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

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

    \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
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  2. #2
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    Quote Originally Posted by jfinkbei View Post

    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

     u=\tan x \implies du=\frac{1}{\cos^2x}
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  3. #3
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    Haha, I knew that... thanks a lot though! could have saved myself some time writing latex had I payed more attention...
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  4. #4
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    Quote Originally Posted by pickslides View Post
    No you're not

     u=\tan x \implies du=\frac{{\color{red}dx}}{\cos^2x}
    ------
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Miss View Post
    ------
    It's usually implied that it's wrt x if dx is not specified.


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

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

    u = \tan (0) = 0

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