# Thread: Another integration by parts problem

1. ## 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
¿Porque?¿

2. 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:

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

$-\frac{1}{2(\alpha+\beta)}\cos(\alpha+\beta)x-\frac{1}{2(\alpha-\beta)}\cos(\alpha-\beta)x+C$

3. 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

4. Hello, Jones!

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

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

5. 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 =/

6. 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.