# Definite integration

1. May 24, 2013

### Dumbledore211

1. The problem statement, all variables and given/known data
Integration of sinx^4

2. Relevant equations
integration by parts

3. The attempt at a solution
I tried to break the function into 2 parts like (sin^3x)(sinx) and solve it using integration by parts but kept getting a function (sin^2x)(cos^2x) over and over again.

2. May 24, 2013

### Cogswell

Does it specifically ask for you to use integration by parts? Maybe using trig identities for $\sin^2 (x)$ will be easier.

If you get a recurring integral, try something like this:

$$\displaystyle \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}] - \int \sin^2 (x) \cos ^2 (x)$$

$$\displaystyle \int \sin^2 (x) \cos ^2 (x) + \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}]$$

$$\displaystyle 2 \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}]$$

$$\displaystyle \int \sin^2 (x) \cos ^2 (x) = \dfrac{[\text{Integral stuff...}]}{2}$$

3. May 24, 2013

### Dumbledore211

It doesn't specifically say anything. I actually tried to use the trig identity sin^2x but got a very messy function to evaluate something like (1/2- 1/2cos2x)^2. I am posting what I actually got by applying integration by parts.

∫sin^3xsinx= sin^3xcosx- 3∫sin^2xcos^2x ............1
=sin^3xcosx - 3[2sin^3xcosx -4∫cos^2xsin^2x]

Let y= ∫sin^3xsinx
from 1 we get, sin^3xcosx -3∫sin^2xcos^2x= sin^3xcosx-6sin^3xcosx+12∫cos^2xsin^2x
→ 6sin^3xcosx= 15∫sin^2xcos^2x
→ 5∫sin^2xcos^2x= 2sin^3xcosx
→ ∫sin^2xcos^2x=2sin^3xcosx/5+C
Please check if there are any reckless mistakes and any step is wrong.

Last edited: May 24, 2013
4. May 24, 2013

### HallsofIvy

First, please note that "$sin x^4$, which is what you wrote, is NOT the same as "$sin^4(x)$ what is what I am sure was intended.

The simplest way to integrate it is to use the fact that $sin^4(x)= (1/2- (1/2)cos(2x))^2$ as you say you started to use but apparently stopped at that point. Continuing, that is equal to $(1/4)- cos(2x)+ (1/4)cos^2(2x)= 1/4- cos(2x)+ (1/4)(1/2+ (1/2)cos(4x)= 3/8- cos(2x)+ 1/8 cos(4x)$.

Now as for your integration by parts, I have no idea what such things as
mean.

5. May 24, 2013

### Ray Vickson

If you know how to do $F(t) = \int \sin^2 (t)\, dt$, then you can write
$$\sin^4 (x) = \sin^2 (x)[1-\cos^2 (x)] = \sin^2 (x) - (\sin(x) \cos(x))^2 = \sin^2 (x) - \frac{1}{4} \sin^2 (2x)$$ and get the solution in terms of $F(x)$ and $F(2x)$.

Last edited: May 24, 2013