Another integration by parts problem

Printable View

August 12th 2009, 02:26 PM
Jones

Another integration by parts problem

Howdy,

Trying to integrate $sin7x~\times~cos8x$

$\text{let}~~u = sin7x, du = 7cos7x$

2 $~sin7x~ \times~cos8x-\int 7cos7x *\frac{sin8x}{8}$

Again..
$\text{Let}~~u = \frac{sin8x}{8}~~ \text{then}~~du=cos8x$

$7cos7\times\frac{sin8x}{8}-\int sin7x*cos8x$

Substitute back into 2

$sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}-\int sin7x*cos8x \bigg]$

$2\int sin7x~\times~cos8x = sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}\bigg]$

$\int sin7x~\times~cos8x = \frac{sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}\bigg]}{2}$

But this is wrong (Crying)
¿Porque?¿
August 12th 2009, 03:03 PM
red_dog

This is the general form of the integral:

$I=\int\sin\alpha x\cos\beta xdx=-\int\left(\frac{\cos\alpha x}{\alpha}\right)'\cos\beta xdx=$

$=-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha}\int\cos\alpha x\sin\beta xdx=$

$=-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha}\int\left(\frac{\sin\alpha x}{\alpha}\right)'\sin\beta xdx=$

$=-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2}\sin\alpha x\sin\beta x+\frac{\beta^2}{\alpha^2}I$

$=\frac{\alpha^2-\beta^2}{\alpha^2}I=-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2}\sin\alpha x\sin\beta x\Rightarrow$

$\Rightarrow I=-\frac{\alpha}{\alpha^2-\beta^2}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2-\beta^2}\sin\alpha x\sin\beta x+C$

Here is another method, without using integration by parts:

$\int\sin\alpha x\cos\beta xdx=\frac{1}{2}\int[\sin(\alpha+\beta)x+\sin(\alpha-\beta)x]dx=$

$-\frac{1}{2(\alpha+\beta)}\cos(\alpha+\beta)x-\frac{1}{2(\alpha-\beta)}\cos(\alpha-\beta)x+C$
August 12th 2009, 03:06 PM
skeeter

parts is too much work ...

$\sin(15x) = \sin(7x+8x) = \sin(7x)\cos(8x) + \cos(7x)\sin(8x) $

$\sin(-x) = \sin(7x-8x) = \sin(7x)\cos(8x) - \cos(7x)\sin(8x) $
----------------------------------------------------------

add up the terms of both equations ...

$\sin(15x) + \sin(-x) = 2\sin(7x)\cos(8x) $

$\sin(7x)\cos(8x) = \frac{1}{2}[\sin(15x) - \sin(x)] $

integrate the RHS
August 12th 2009, 03:50 PM
Soroban

Hello, Jones!

skeeter is right . . . "by parts" is a lot of work.
Here it is . . . in case you ever get that desperate.

Quote:

Integrate: . $I \;=\;\int \sin7x\cos8x\,dx$

By parts: . $\begin{array}{cccccccc}u &=&\sin7x & & dv &=& \cos8x\,dx \\ du &=& 7\cos7x\,dx & & v&=&\frac{1}{8}sin8x \end{array}$

. . We have: . $I \;=\;\tfrac{1}{8}\sin7x\sin8x - \tfrac{7}{8}\!\!\int\sin8x\cos7x\,dx$

By parts: . $\begin{array}{cccccccc}u &=& \cos7x && dv &=& \sin8x\,dx \\ du &=& \text{-}7\sin7x\,dx & & v &=& \text{-}\frac{1}{8}\cos8x \end{array}$

. . We have: . $I \;=\;\tfrac{1}{8}\sin7x\sin9x - \tfrac{7}{8}\bigg[\text{-}\tfrac{1}{8}\cos7x\cos8x - \tfrac{7}{8}\int\!\!\sin7x\cos8x\,dx\bigg]$

$\text{Hence: }\;I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}\cos7x\cos8x + \tfrac{49}{64}\!\!\underbrace{\int\sin7x\cos8x\,dx }_{\text{This is }I}+C$

We have: . $I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}\cos7x\cos8x + \tfrac{49}{64}I + C$

. . Then: . $\tfrac{15}{64}I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}cos7x\cos8x + C$

. . Multiply by $\tfrac{64}{15}\!:\quad I \;=\;\tfrac{8}{15}\sin7x\sin8x + \tfrac{7}{15}\cos7x\cos8x + C$

Therefore: . $\int\sin7x\cos8x\,dx \;=\;\frac{8}{15}\sin7x\sin8x + \frac{7}{15}\cos7x\cos8x + C$
August 13th 2009, 12:29 AM
Jones

Thanks,

How do you know when to use integration by parts and when not to use it?

I thought you always had to use it if you had a product of two functions =/
August 13th 2009, 01:18 AM
Prove It

Quote:

Originally Posted by Jones

Thanks,

How do you know when to use integration by parts and when not to use it?

I thought you always had to use it if you had a product of two functions =/

You use integration by parts if two situations hold...

1. One of the functions in the product should be easy to differentiate and one should be easy to integrate.

2. By integrating and differentiating the correct functions, the new integral is reduced to something easier to integrate.