# Tricky trig substitution

1. Apr 24, 2005

### p53ud0 dr34m5

im hoping i worked this out right; its long:
$$\int x(81-x^2)^{5/2}dx$$
the integral contains $a^2-x^2$, so i set $x=asin\theta$. that would make $x=9sin\theta$ and $dx=9cos\theta d\theta$:
$$\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta d\theta$$
the integral now contains $1-sin^2\theta =cos^2\theta$
$$\int 9sin\theta(9cos\theta)^59cos\theta d\theta$$
i used u-sub by setting $u=9cos\theta$ and $du=-9sin\theta d\theta$.
$$-\int u^6 du=-\frac{u^7}{7}+C$$
i plugged my u back in:
$$-\frac{(9cos\theta)^7}{7}+C$$
then, i drew my little triangle.
$$cos\theta=\frac{\sqrt{81-x^2}}{9}$$
i plugged that into the $cos\theta$ and simplified and came out with:
$$-\frac{(\sqrt{81-x^2})^7}{7}+C$$

2. Apr 24, 2005

### Muzza

You could also try the substitution u = 81 - x^2, which should be significantly easier.

3. Apr 24, 2005

### Theelectricchild

I punched it into mathematica and it produced the same as yours, save for a 7/2 exponent rather than 7.

4. Apr 24, 2005

### p53ud0 dr34m5

7/2 exponent is like my square root to the 7th
$$-\frac{(\sqrt{81-x^2})^2}{7}=-\frac{(81-x^2)^{7/2)}{7}$$
also, we had to use trig substitution, so that left out the easy 81-x^2 sub

5. Apr 24, 2005

### Hippo

It doesn't have to be.

You shouldn't have to use a trig or u substitution. Inspection isn't too hard with something like this.

$$\frac{(2)}{(-2)(7)}\int (-2x)(\frac{7}{2})(81-x^2)^{5/2}dx$$
$$= -\frac{1}{7} (\sqrt {81-x^2})^7$$

Last edited by a moderator: Apr 24, 2005
6. Apr 24, 2005

### Theelectricchild

*Cough* Table *Cough*

Excuse me... something in my throat--- tables are great to have around so you barely have to think.

7. Apr 24, 2005

### Hippo

Guess having someone else do it is the easiest way.