# Indefinite integral of sin^4xdx

1. Feb 20, 2006

### Captain Zappo

Hi everyone. I'm having some trouble evaluating the following integral

$$\int{sin^4xdx}$$

First let me start off by showing what I did.

$$= \int{(sin^2x)(sin^2x)dx}$$

$$=\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx$$

$$=\int{(\frac{1}{4}-cos(2x)+\frac{1}{4}cos^2(2x))dx$$

$$=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{4}\int{(\frac{1}{2}+\frac{1}{2}cos(4x))dx$$

$$=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{8}\int{cos(4x)}dx$$

$$=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{32}sin(4x)+C$$

I'm not sure if I have the right answer or not, so could someone please check and see if I did anything illegal. If so, can you please correct me, or even show me a different method entirely.

Thanks,
-Zach

Last edited: Feb 20, 2006
2. Feb 20, 2006

### quasar987

$$=\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx$$

$$=\int{(\frac{1}{4}-\frac{1}{2}cos(2x)+\frac{1}{4}cos^2(2x))dx$$

no?

3. Feb 20, 2006

### Captain Zappo

Thank you!!

4. Jun 17, 2008

### c-murda

why is the second cos squared?

what is next step after that? im stuck here too...

Last edited: Jun 17, 2008
5. Jun 17, 2008

### cakesama

expand the expression in the integral, then you get the last term being cos sqaured

6. Jun 17, 2008

### c-murda

7. Jun 17, 2008

### cakesama

so you expand to get this

$$=\int{(\frac{1}{4}-\frac{1}{2}cos(2x)+\frac{1}{4}cos^2(2x))dx$$

then its an integral of 3 terms, so you integrate the 3 terms separately
did that help?

8. Jun 17, 2008

### Dick

There's a trig formula for cos^2(2x) in terms of cos(4x). Just look at what Captain Zappo tried to do, in spite of fluffing a coefficient.

9. Jun 18, 2008

### c-murda

∫sin^4(x)

[sin^2(x)]^2

[(1 -cos(2x)/2)]^2

1/4[1 + cos^2(2x) - 2cos(2x)]

1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]

1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)

(3/8) - (1/2)cos(2x) + (1/8)cos(4x)

so ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx

∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c

is this right?

10. Jun 18, 2008

### Dick

Looks fine to me.

11. Jun 18, 2008

### rootX

http://integrals.wolfram.com/index.jsp?expr=(sin(x))^4&random=false

You should use this (for other questions)

12. Jun 18, 2008

### m_s_a

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c
correct