# Another integration by parts problem

• Aug 12th 2009, 02:26 PM
Jones
Another integration by parts problem
Howdy,

Trying to integrate $\displaystyle sin7x~\times~cos8x$

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

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

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

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

Substitute back into 2

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

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

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

But this is wrong (Crying)
¿Porque?¿
• Aug 12th 2009, 03:03 PM
red_dog
This is the general form of the integral:

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

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

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

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

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

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

Here is another method, without using integration by parts:

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

$\displaystyle -\frac{1}{2(\alpha+\beta)}\cos(\alpha+\beta)x-\frac{1}{2(\alpha-\beta)}\cos(\alpha-\beta)x+C$
• Aug 12th 2009, 03:06 PM
skeeter
parts is too much work ...

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

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

add up the terms of both equations ...

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

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

integrate the RHS
• Aug 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: .$\displaystyle I \;=\;\int \sin7x\cos8x\,dx$

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

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

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

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

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

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

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

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

Therefore: .$\displaystyle \int\sin7x\cos8x\,dx \;=\;\frac{8}{15}\sin7x\sin8x + \frac{7}{15}\cos7x\cos8x + C$

• Aug 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 =/
• Aug 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.