integrate e^x coshx dx

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- Feb 8th 2010, 02:35 AMconfusedgirlAnother problem on Integration by parts
integrate e^x coshx dx

- Feb 8th 2010, 03:43 AMjiboom
- Feb 8th 2010, 04:01 AMHallsofIvy
As jiboom suggested, the simplest thing to do is to use the definition of cosh(x) in terms of $\displaystyle e^x$. What are you using as the definition of cosh(x)?

- Feb 8th 2010, 04:03 AMconfusedgirl
(e^x + e^-x)/2

- Feb 8th 2010, 04:14 AMconfusedgirl
- Feb 8th 2010, 04:16 AMArchie Meade
$\displaystyle Cosh(x)=\frac{e^x+e^{-x}}{2}$

$\displaystyle e^xCosh(x)=\frac{e^xe^x+e^xe^{-x}}{2}=\frac{e^{2x}+e^0}{2}$

$\displaystyle =\frac{e^{2x}+1}{2}$

$\displaystyle \frac{1}{2}\int{e^{2x}}dx+\int{\frac{1}{2}}dx=\fra c{1}{2}\int{e^xe^x}dx+\frac{x}{2}+C$

To integrate by parts

Let $\displaystyle \int{e^xe^x}dx=I$

$\displaystyle u=e^x\ \Rightarrow\ du=e^xdx$

$\displaystyle dv=e^xdx\ \Rightarrow\ v=e^x$

$\displaystyle uv-\int{v}du=e^xe^x-\int{e^xe^x}dx$

$\displaystyle I=e^{2x}-I$

$\displaystyle 2I=e^{2x}$

$\displaystyle I=\frac{e^{2x}}{2}+C$

Therefore

$\displaystyle \int{e^xCosh(x)}dx=\frac{1}{4}e^{2x}+\frac{1}{2}x+ C$ - Feb 8th 2010, 08:50 AMjiboom