# Simple derivative problem

1. Feb 4, 2004

### calculateme

I am taking basic calculus, and have just got to integration. Can someone please tell me how to find the antiderivative of (sin(x))^4?

2. Feb 4, 2004

### matt grime

Looks like home work. Replace a power with a multiple - you can do sin^2 using cos 2, so this is no harder.

3. Feb 4, 2004

### calculateme

Sorry, but I still don't understand. How do you find the antiderivative of (sin(x))^2? Could you explain it to me please?

4. Feb 4, 2004

### NateTG

Do you know about the chain rule?
do you know the derivatives of $$\sin x$$ and $$x^2$$?

5. Feb 5, 2004

### calculateme

Yes, but how are they going to help me find the antiderivative of $$sin^4x$$?

6. Feb 5, 2004

### himanshu121

Try to Reduce the power of sin4x by
2sin2x=1-cos2x.

therefore
4sin4x=(1-cos2x)2

i.e 1+cos22x-2cos2x
Again use

2cos22x = 1+cos24x

Simplifying u will obtain

$$\sin^4x = \frac{3}{8} +\frac{cos4x}{8}-\frac{cos2x}{2}$$

Hope this will help u

7. Feb 5, 2004

### Integral

Staff Emeritus
Himanshu

I think you have a typo

Should be

8. Feb 5, 2004

### himanshu121

Ya typo is there it is

2cos22x = 1+cos4x

error is regretted

9. Feb 5, 2004

### calculateme

Himanshu

Thanks a lot, I understand now.