Another problem on Integration by parts

**confusedgirl** · February 8th 2010, 02:35 AM

integrate e^x coshx dx

**jiboom** · February 8th 2010, 03:43 AM

Originally Posted by confusedgirl

integrate e^x coshx dx

if only you could write this as the integral of e^x[something involving e^x and e^(-x)]

This bit was only good to show 0=0 so no piont keeping it

**HallsofIvy** · February 8th 2010, 04:01 AM

As jiboom suggested, the simplest thing to do is to use the definition of cosh(x) in terms of $e^x$ . What are you using as the definition of cosh(x)?

**confusedgirl** · February 8th 2010, 04:03 AM

(e^x + e^-x)/2

**confusedgirl** · February 8th 2010, 04:14 AM

Originally Posted by jiboom

if only you could write this as the integral of e^x[something involving e^x and e^(-x)]

or,if you have to do it by parts, just use parts twice. If we call the integral I., when you do parts twice you will end up with

I=something+kI whrer something and k are found from doing parts

which we can arrange as

(1-k)I=something

and so I can be found.

if i integrate by parts 2 times
the result will be I = 0/2 which is equal to zero

**Archie Meade** · February 8th 2010, 04:16 AM

Originally Posted by confusedgirl

integrate e^x coshx dx

$Cosh(x)=\frac{e^x+e^{-x}}{2}$

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

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

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

To integrate by parts

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

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

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

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

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

$2I=e^{2x}$

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

Therefore

$\int{e^xCosh(x)}dx=\frac{1}{4}e^{2x}+\frac{1}{2}x+ C$

**jiboom** · February 8th 2010, 08:50 AM

Originally Posted by confusedgirl

if i integrate by parts 2 times
the result will be I = 0/2 which is equal to zero

i only added that as an edit as you wanted parts. I have never done that sort of integral by parts, i would use my first (and useful comment).

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