# Trigonometric Substitution

1. Apr 7, 2005

### amcavoy

$$\int x^3\sqrt{4-9x^2}dx$$

I tried to use $$x=\frac{2}{3}\cos{(x)}$$ but it just left me with $$\int \sin^3{(x)}\cos^2{(x)}dx$$

Any suggestions?

2. Apr 7, 2005

### dextercioby

It's the oher way around.U missed a numerical factor (containing an essential minus)...U need to substitute

$$x=\frac{2}{3}\sin u$$

to get the powers of sine & cosine that u have there.U'd still miss the numerical factor and but at least you'd have gotten rid of the "-"...

Let's assume this one (it's your transformed integral up to a numerical factor).

$$\int \sin^{3}u \cos^{2} u \ du$$...

Use the fundamental identity of circular trigonometry to write it

$$\int \sin^{3}u \ du -\int \sin^{5}u \ du$$

The first can be written as

$$-\int (1-\cos^{2}u) \ d(\cos u)$$

The second (the integral,without the "-" preceding it).

$$-\int \left(1-\cos^{2}u\right)^{2} \ d(\cos u)$$

,which are very easy to integrate...

Daniel.

Last edited: Apr 7, 2005
3. Apr 7, 2005

### amcavoy

$$\int{\sin^3{\theta}}d\theta - \int{\sin^5{\theta}}d\theta$$

I know how to integrate the first, but the second I am having a bit more trouble with.

Is this true?: $$\int{\sin^5{\theta}}d\theta = -\int{(1-\cos{\theta})^2}d\cos{\theta}$$

Thanks again.

4. Apr 7, 2005

### dextercioby

I told u what it is equal to.

$$\int \sin^{5}x \ dx=-\int \left(1-\cos^{2}x\right)^{2} d(\cos x)$$

Daniel.

5. Apr 7, 2005

### amcavoy

I'm sorry, what I meant was:

What is the best way to integrate it? I haven't seen one with the square before, so I'm not sure...

6. Apr 7, 2005

### HallsofIvy

Staff Emeritus
If you are going to use "trigonometric substitution" you should learn to integrate trig functions first!

Any time you have a an odd power of sin or cos, factor out one of then to go with the "dx", then convert the even power using sin2x+ cos2x= 1.

For example, seeing $$\int sin^5 x dx$$, I thinkof it as $$\int sin^4 x (sin x dx)= \int (sin^2 x)^2 (sin x dx)= \int (1- cos^2 x)(sin x dx)$$. Now let u= cos(x) so that du= -sin x dx and that becomes $$\int (1-u^2)^2 du$$.

If you have only even powers of sin and cos, you can't do that- you have to use the trig identities $$cos^2(x)= \frac{1}{2}[1+ cos(2x)]$$ and $$sin^2(x)= \frac{1}{2}[1- cos(2x)]$$ to reduce the powers.