# Thread: Intergration

1. ## Intergration

This problem I cannot seem to solve,I have the answer but i want to know how to get to it, i do not know if i used the table or do i integrate it by parts? Thank You!!
The problem:
Integrate[Sin[3 x] Cos[5 x], x]

and this is the answer:

8 (Cos[x]) ^(2)- Cos[8 x]
-------------------------
16

2. Make

$\displaystyle \int \sin(3x)\cos(5x)~dx$

into

$\displaystyle \int \sin(4x-x)\cos(4x+x)~dx$

and using Sum-Difference formulas from here

Table of Trigonometric Identities

$\displaystyle \int \sin(4x-x)\cos(4x+x)~dx=$ $\displaystyle \int (\sin(4x)\sin(x)-\cos(4x)\cos(x))(\cos(4x)\cos(x)-\sin(4x)\sin(x))~dx$

now simplify and integrate.

3. Originally Posted by pickslides
Make

$\displaystyle \int \sin(3x)\cos(5x)~dx$

into

$\displaystyle \int \sin(4x-x)\cos(4x+x)~dx$

and using Sum-Difference formulas from here

Table of Trigonometric Identities

$\displaystyle \int \sin(4x-x)\cos(4x+x)~dx=$ $\displaystyle \int (\sin(4x)\sin(x)-\cos(4x)\cos(x))(\cos(4x)\cos(x)-\sin(4x)\sin(x))~dx$

now simplify and integrate.
at the end when you plug back in the trig identities, the sum-difference,from sin(4x-x) is not right

4. $\displaystyle \sin(4x-x)=\sin(4x)\cos(x)-\cos(4x)\sin(x)$

5. Originally Posted by flutterby
This problem I cannot seem to solve,I have the answer but i want to know how to get to it, i do not know if i used the table or do i integrate it by parts? Thank You!!
The problem:
Integrate[Sin[3 x] Cos[5 x], x]

and this is the answer:

8 (Cos[x]) ^(2)- Cos[8 x]
-------------------------
16
You can also do this using the "cyclic method". (That's not it's official name; that's just what I call it.)

Let $\displaystyle I=\int\sin3x\cos5x\,dx$

$\displaystyle u=\sin3x$ and $\displaystyle dv=\cos5x\,dx$
$\displaystyle du=3\cos3x\,dx$ and $\displaystyle v=\frac{1}{5}\sin5x$

The integration by parts formula gives:

$\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\int\cos3x\sin5x\,dx$

$\displaystyle u=\cos3x$ and $\displaystyle dv=\sin5x\,dx$
$\displaystyle du=-3\sin3x\,dx$ and $\displaystyle v=-\frac{1}{5}\cos5x$

Now we have:

$\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\int\cos3x\sin5x\,dx$ $\displaystyle =\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\left[-\frac{1}{5}\cos3x\cos5x-\frac{3}{5}\int\sin3x\cos5x\,dx\right]$

Note that that last integral is just $\displaystyle I$, so we have:

$\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\left[-\frac{1}{5}\cos3x\cos5x-\frac{3}{5}I\right]$

Solve for $\displaystyle I$. (You should also check my math; make sure I didn't make any sign errors and whatnot.)

Remark: To get the answer into the format given to you in the answer key, you may need to use some trig identities.