# Integrate (sin x^4)

1. Jan 2, 2006

### teng125

may i know how to integ (sin x^4) ??

the answer is 1/32(12x - 8sin 2x + sin4x)

2. Jan 2, 2006

### VietDao29

Again, you can use Power-reduction formulas. Then use some Product-to-sum identities, your goal is convert that sin(x) to the power of 4 into some sine or cosine functions to the power of 1.
Now let's first split sin4x into (sin2x sin2x). Can you go from here?

3. Jan 2, 2006

### teng125

i try to subs using cos2x=1-s(sinx)^2 but can't get

4. Jan 2, 2006

### VietDao29

So you have:
cos(2x) = cos2x - sin2x = 2cos2x - 1 = 1 - 2sin2x.
From there, rearrange them a bit, you will have:
$$\cos ^ 2 x = \frac{1 + \cos(2x)}{2} \quad \mbox{and} \quad \sin ^ 2 x = \frac{1 - \cos(2x)}{2}$$
These are call Power-reduction formulas.
So we will now use $$\sin ^ 2 x = \frac{1 - \cos(2x)}{2}$$.
$$\int \sin ^ 4 x dx = \int (\sin ^ 2 x) ^ 2 dx = \int \left( \frac{1 - \cos(2x)}{2} \right) ^ 2 dx = \frac{1}{4} \int ( 1 - \cos(2x) ) ^ 2 dx$$
$$= \frac{1}{4} \int ( 1 - 2 \cos(2x) + \cos ^ 2 (2x)) dx$$.
Now again use the Power-reduction formulas for cos2(2x).
Can you go from here?

5. Jan 2, 2006

### teng125

ya,that's where i got stuck because i don't know how to get the sin4x.how to obtain 1/32 sin4x??

6. Jan 2, 2006

### VietDao29

Did I tell you to use the Power-reduction formulas for cos2(2x). It's the last line of my above post (namely, the #4 post of this thread).
Since you have:
$$\cos ^ 2 x = \frac{1 + \cos(2x)}{2}$$, so that means:
$$\cos ^ 2 (2x) = \frac{1 + \cos(2 \times (2x))}{2} = \frac{1 + \cos(4x)}{2}$$.
Can you go from here?

7. Jan 2, 2006

### teng125

oh........okok i saw it.....thanx very much