# Sine and cos integration

1. Oct 14, 2006

### meee

I need to find $$\int \cos^2(x)\sin^7(x)dx$$

I'm not sure what substitution to make

2. Oct 14, 2006

### Office_Shredder

Staff Emeritus
If you start by turning $$\cos^2 (x)$$ into $$1 - \sin^2(x)$$, you get an integral that can be done by a reduction formula.

3. Oct 14, 2006

### FunkyDwarf

Or you can do intergration by parts but the above way is prob quicker

4. Oct 14, 2006

### meee

ok so now i got

$$\int (1-\sin^2(x))\sin^7(x)dx$$

and then what?

$$\int \sin^7(x)-\sin^9(x) dx$$ ?? and now im stuck :(

and then $$\int (1-\cos^2(x))^3\sin(x) - (1-\cos^2(x))^4\sin(x)) dx$$

ahh

Last edited: Oct 14, 2006
5. Oct 14, 2006

### Office_Shredder

Staff Emeritus
Try doing $$\int \sin^n(x)dx$$ by integration by parts, to get a formula in terms of an integral of sin to a lower power

6. Oct 14, 2006

### meee

sorry, im not sure what integration by parts is?

7. Oct 14, 2006

### Office_Shredder

Staff Emeritus
integration by parts:

$$\int udv = uv - \int vdu$$

basically, if you have an integral that is the product of a function u, and the derivative of another function v, it equals u*v - the integral of v times the derivative of u. Here's an example:

$$\int x*e^xdx$$ Say u=x, and dv=exdx. Then v=ex, and du=dx. So:

$$\int x*e^xdx = x*e^x - \int e^xdx = x*e^x - e^x$$

8. Oct 14, 2006

### neutrino

Learning another method to integrate is good, but if meee is required to do this particular problem by just using sub. methods, then this is probably a way.

Rewrite the integral as...

$$\int{\cos^2{x}\sin^6{x}\sin{x}dx}$$

Convert the $$\sin^6{x}$$ in terms of cos, and the do the substitution u = cos(x).

9. Oct 14, 2006

### meee

ok thanks guys...

Office_Shredder thanks for that.. but i think integration by parts is beyond my current course.

neutrino, i tried what you said, but it looks kinda weird...

$$\int{\cos^2{x}\sin^6{x}\sin{x}dx}$$

$$= \int{\cos^2{x}(1-\cos^2{x})^3\sin{x}dx}$$

let u = $$\cos{x}$$
$$\frac{du}{dx} = -\sin{x}$$

$$= -\int{u^2(1-u^2)^3 du}$$
$$= -\int{u^2(1-3u^2+3u^4-u^6) du}$$
$$= -\int{u^2-3u^4+3u^6-u^8)du}$$
$$= -(\frac{1}{3}u^3-\frac{3}{5}u^5+\frac{3}{7}u^7-\frac{1}{9}u^9)$$

and then sub u=cos(x) back in

is that right?

Last edited: Oct 14, 2006
10. Oct 14, 2006

that is correct

11. Oct 14, 2006

### meee

thanks :D cool

12. Oct 14, 2006

### neutrino

Yes.

Btw, I don't think you'll have to wait till university to find more about integration by parts. You'll definitely meet it at the school level.

13. Oct 14, 2006

### meee

oh reali?
ive finished my year 12 course (exams in 2 weeks), and havent seen it in school or outside school lectures.

maybe it will be useful for me to learn it.

14. Oct 14, 2006

### neutrino

Okay, shouldn't have generalised. I was under the assumption that every school, in every part of the world taught this.

It most certainly will.

15. Oct 14, 2006

### meee

ok, thanks

and lol im in australia our schools are probly weird

16. Oct 27, 2006

### CDevo69

My school which is in Australia are doing it now. Its fairly easy to learn.