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Math Help - Integral... where is my error

  1. #1
    MHF Contributor arbolis's Avatar
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    Integral... where is my error

    I've done a pretty similar one, but for this one I don't get what I should get. I post the integral and my steps : \omega=\int_0^{\frac{\pi}{4}} \frac{dx}{1+\sin (x)}.
    Let x=2\arctan (u), dx= \frac{2}{1+u^2}du.
    \Rightarrow \omega=\int \frac{\frac{2}{1+u^2}}{1+\sin (2\arctan(u))}du. Let \theta=\arctan (u). \Rightarrow \sin (2\arctan (u))=\sin (2\theta)=2\cos (\theta)\sin (\theta). But as \cos (\theta)=\frac{1}{\sqrt{1+u^2}} and \sin (\theta)=\frac{u}{\sqrt{1+u^2}}, we have that \omega= \int \frac{\frac{2}{1+u^2}}{\frac{2u}{1+u^2}}du=\int \frac{du}{u}=\ln |u| +C, backsub I get \ln |\tan \left( \frac{x}{2}\right)|+C. Putting the limits I finally get that \omega is worth \ln|\tan \left( \frac{\pi}{8}\right) |+C since  \tan(0)=0. But the answer should be 2-\sqrt 2 according to Mathematica...
    I didn't see my error even when typing all this in Latex. Note that for simplicity I didn't take in count the limits of the integral, so it's not mathematically correct to say that \omega is worth all the steps I did before to get to the last value (from which it should be equal to, but it seems it isn't).
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    Quote Originally Posted by arbolis View Post
    I've done a pretty similar one, but for this one I don't get what I should get. I post the integral and my steps : \omega=\int_0^{\frac{\pi}{4}} \frac{dx}{1+\sin (x)}.
    .
    \frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} = \frac{1-\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos^2 x}.
    Can you finish it from there?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Can you finish it from there?
    Thanks a lot, I think yes. I must go to sleep now so I will try tomorrow. But to my eyes I have to do an integration by parts for \int \sec^2(x)dx and a u-sub for -\int \frac{\sin(x)}{cos^2(x)}dx. I'd be grateful also if someone point an error of mine when I did my trig-sub since I don't see where I make an error.
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    Quote Originally Posted by arbolis View Post
    But to my eyes I have to do an integration by parts for \int \sec^2(x)dx .
    This is simply \tan x + k.
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by arbolis View Post
    I'd be grateful also if someone point an error of mine when I did my trig-sub since I don't see where I make an error.

    you forgot that there was a 1 in the denominator of the original equation..

    your final integral should have been

    \omega = \int\frac{du}{(1+u)^2}
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