# trig sub

• Mar 26th 2009, 12:14 PM
qzno
trig sub
Find the following integral:

$\displaystyle \int e^x \sqrt{1-e^{2x}}$
• Mar 26th 2009, 12:23 PM
Chris L T521
Quote:

Originally Posted by qzno
Find the following integral:

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

Let $\displaystyle e^x=\sin\theta$

Thus, $\displaystyle e^x\,dx=\cos\theta\,d\theta$

The integral becomes $\displaystyle \int\cos\theta\sqrt{1-\sin^2\theta}\,d\theta=\int\cos^2\theta\,d\theta$

Now recall that $\displaystyle \cos^2\theta=\frac{1+\cos\!\left(2\theta\right)}{2 }$

Can you continue the problem?
• Mar 26th 2009, 12:31 PM
qzno
dont you mean that:

$\displaystyle$
$\displaystyle \int\sin\theta\sqrt{1-\sin^2\theta}\,d\theta$

because

$\displaystyle e^x = sin\theta$
• Mar 26th 2009, 12:33 PM
qzno
ooh my bad, you did some cancelling there i see haha
• Mar 26th 2009, 12:37 PM
qzno
I got the answer to be:

$\displaystyle sin^{-1} (e^x) - \frac{e^x \sqrt{1-e^{2x}}}{2} + C$

Is this correct?
• Mar 26th 2009, 12:45 PM
Chris L T521
Quote:

Originally Posted by qzno
I got the answer to be:

$\displaystyle sin^{-1} (e^x) - \frac{e^x \sqrt{1-e^{2x}}}{2} + C$

Is this correct?

Close...

When you integrate $\displaystyle \tfrac{1}{2}\int 1+\cos\left(2\theta\right)\,d\theta$, you get $\displaystyle \tfrac{1}{2}\theta+\tfrac{1}{4}\sin\left(2\theta\r ight)+C=\tfrac{1}{2}\theta+\tfrac{1}{2}\sin\theta\ cos\theta+C$, where $\displaystyle \sin\left(2\theta\right)=2\sin\theta\cos\theta$

Now substituting back in, we have $\displaystyle \theta=\sin^{-1}\left(e^x\right)$, $\displaystyle \sin\theta=e^x$, and $\displaystyle \cos\theta=\sqrt{1-e^{2x}}$

Therefore, $\displaystyle \tfrac{1}{2}\theta+\tfrac{1}{2}\sin\theta\cos\thet a+C=\boxed{\tfrac{1}{2}\left(\sin^{-1}\left(e^x\right)+e^x\sqrt{1-e^{2x}}\right)+C}$
• Mar 26th 2009, 12:55 PM
qzno
excellent thank you : )