1. ## Intergration question

intergrate:

1) cos2xsin3x dx
2) 1+x ((x+Inx))^0.5
x

3) got this question sitting in front of me and not sure how to get my head about it sounds simple but aint got a scooby doo:

'Find the volume generated by rotating the area bounded by y=x^2 and y=4 around the line y=-1

ty x

2. Originally Posted by Amanda-UK
intergrate:

1) cos2xsin3x dx
$\int cos(2x)sin(3x)dx$ or $\int cos^2(x) sin^3(x) dx$?

-Dan

3. first one

amanda x

4. $\int cos(2x)sin(3x)dx$

The simplest attack I can think of is to use:
$cos(2x) = 1 - 2sin^2(x)$
$sin(3x) = -4sin^3(x) + 3sin^2(x)$

So:
$\int cos(2x)sin(3x)dx = \int (1 - 2sin^2(x))(-4sin^3(x) + 3sin^2(x))dx$

$= \int (8sin^5(x) - 6sin^4(x) -4sin^3(x) + 3sin^2(x)) dx$

Then use
$\int sin^n(x) dx = -\frac{sin^{n-1}(x)cos(x)}{n} + \frac{n-1}{n}\int sin^{n - 2}(x) dx$

-Dan

5. Originally Posted by topsquark
$\int cos(2x)sin(3x)dx$

The simplest attack I can think of is to use:
$cos(2x) = 1 - 2sin^2(x)$
$sin(3x) = -4sin^3(x) + 3sin^2(x)$

So:
$\int cos(2x)sin(3x)dx = \int (1 - 2sin^2(x))(-4sin^3(x) + 3sin^2(x))dx$

$= \int (8sin^5(x) - 6sin^4(x) -4sin^3(x) + 3sin^2(x)) dx$

Then use
$\int sin^n(x) dx = -\frac{sin^{n-1}(x)cos(x)}{n} + \frac{n-1}{n}\int sin^{n - 2}(x) dx$

-Dan
do i use the last bit to substitute the bit before? after i do this what will i need to do next? amanda xx

6. Originally Posted by Amanda-UK
do i use the last bit to substitute the bit before? after i do this what will i need to do next? amanda xx
Use the recursion relation to reduce the $sin^5(x)$ integral to a $sin^3(x)$ integral, then to a $sin(x)$ integral. etc.

-Dan

7. i have never used that method is it possible to do it by using intergration by parts? amanda x

8. Originally Posted by topsquark
$\int cos(2x)sin(3x)dx$

The simplest attack I can think of is to use:
$cos(2x) = 1 - 2sin^2(x)$
$sin(3x) = -4sin^3(x) + 3sin^2(x)$
Hello Dan

Did you try to use $\sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin( \alpha-\beta)}2$ ?

9. For the solid of revolution, you can washers or shells.

Washers:

${\pi}\int_{-2}^{2}[(-1-4)^{2}-(-1-x^{2})^{2}]dx$

Shells:

$4{\pi}\int_{0}^{4}[(y+1)\sqrt{y}]dy$

The diagram was supposed to be animated, but I had to resize and then it wouldn't work.

10. Originally Posted by Krizalid
Hello Dan

Did you try to use $\sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin( \alpha-\beta)}2$ ?
You know, I recently showed that one to another poster... Ah well!

-Dan