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Math Help - integration

  1. #1
    Member i_zz_y_ill's Avatar
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    integration

    I was doing this equation \int e^x\aqrt{1-e^2x}dx and I used the substitution e^x=sinu which gave me

    \frac{sin(arcsin(e^x))(cos(arcsin(e^x))}{2}+\frac{  arcsin(e^x)}{2}+C
    which simplifies to \frac{e^x\sqrt{1-exp(2x)}}{2}+\frac{arcsin(e^x)}{2}+C but i cant see how the first term of this was obtained from the first line here. If some one could help youd be od great help!!!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    I was doing this equation \int e^x\aqrt{1-e^2x}dx and I used the substitution e^x=sinu which gave me

    \frac{sin(arcsin(e^x))(cos(arcsin(e^x))}{2}+\frac{  arcsin(e^x)}{2}+C
    which apparantly simplifies to \frac{e^x\sqrt{1-exp(2x)}}{2}+\frac{arcsin(e^x)}{2}+C but i cant see how in my previous thread how this was done If some one could help youd be od great help!!!
    \int\sqrt{1-u^2}du

    If we let u=\sin(z)

    Then we get

    \int\cos^2(z)dz

    =\frac{1}{2}\int\bigg[1+\cos(2z)\bigg]dz

    =\frac{1}{2}\bigg[z+\frac{1}{2}\sin(2z)\bigg]+C(1)

    So now seeing that

    \sin(2z)=2\sin(z)\cos(z)=2\sin(z)\sqrt{1-\sin^2(z)}

    We can rewrite (1) as

    \frac{1}{2}\bigg[z+\sin(z)\sqrt{1-\sin^2(z)}\bigg]+C

    Now backsubbing \arcsin(u)=z

    Gives us

    \frac{1}{2}\bigg[\arcsin(u)+\sin\left(\arcsin(u)\right)\sqrt{1-\sin^2\left(\arcsin(u)\right)}\bigg]+C

    Now you really should know that

    \sin\left(\arcsin(z)\right)=z

    So using that we may further rewrite (1) as

    \frac{1}{2}\bigg[\arcsin(u)+u\sqrt{1-u^2}\bigg]+C(2)

    So now you asked

    \int{e^x\sqrt{1-e^{2x}}dx}

    Now by the sub m=e^x

    We obtain

    \int\sqrt{1-m^2}dm

    So now using (2) we see that this is equivalent to

    \frac{1}{2}\bigg[\arcsin(m)+m\sqrt{1-m^2}\bigg]+C

    Now backsubbing finally gives

    \boxed{\int{e^x\sqrt{1-e^{2x}}}dx=\frac{1}{2}\bigg[\arcsin\left(e^x\right)+e^x\sqrt{1-e^{2x}}\bigg]+C}

    That is about as drawn out and forward as I can be. If you still dont get it, I am not sure I can help
    Last edited by Mathstud28; November 5th 2008 at 04:14 PM. Reason: typo
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