Finding the Antiderivative of sin^4(x) in Basic Calculus

[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?

 

[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.

 

[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?

 

[NateTG]

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

 

[calculateme]

Originally posted by NateTG
Do you know about the chain rule?
do you know the derivatives of \sin x and x^2?

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

 

[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

 

[Integral]

Himanshu

I think you have a typo

2cos²2x = 1+cos²4x

Should be

2cos²2x = 1+cos4x

 

[himanshu121]

Ya typo is there it is

2cos²2x = 1+cos4x

error is regretted

 

[calculateme]

Himanshu

Thanks a lot, I understand now.