# Calculus II Trig Sub

1. Mar 4, 2009

### pumpkin88

I'm very unsure on how to do this equation...

$$\int$$x$$^{3}$$$$\sqrt{}x^{2}+9$$

i do know that
if you...
let x = 3sec $$\vartheta$$
dx = 3 sec $$\vartheta$$ tan $$\vartheta$$ du

2. Mar 4, 2009

### lubuntu

sqrt of x^2? you sure you wrote that right?

3. Mar 4, 2009

### lubuntu

you wouldn't substitute in a secant, a tangent would do much better, when you sub in you have to figure out what you should put in the place of that x^3 too!

4. Mar 4, 2009

### pumpkin88

yeah. its the integral of x^3 times the sqrt of x^2 + 9

5. Mar 4, 2009

### pumpkin88

so would it be 3tan^3 theta times the sq rt of 3tan^2theta +9?

6. Mar 4, 2009

### lubuntu

and you need d(theta) in there too, plus remember you have to raise the entire term to the power

7. Mar 4, 2009

### pumpkin88

would d(theta) be 3sec(theta)tan(theta) ?

8. Mar 4, 2009

### carlodelmundo

Remember your letting x = 3tan (theta).

x^3 means 9tan^3 (theta) and the other term becomes sqrt [ 9tan^2 (theta) + 9 ]

Factor out a square root of 9 out of the square root and use the identity: 1 + tan^2 x = sec^2 x to "simplify" your problem.

You should be left with an integration of a sec to the third and a tangent to the third.

9. Mar 4, 2009

### carlodelmundo

if you let x = 3tan (theta)

what is the derivative in terms of d (theta) ?

10. Mar 4, 2009

### pumpkin88

is this correct so far?

9 $$\int$$ tan$$^{3}$$ $$\theta$$ $$\sqrt{$$ sec $$^{2}$$ $$\theta$$

11. Mar 4, 2009

### carlodelmundo

Close.

Like lumbar said, you forgot the d(Theta).

if you let x = 3tan(Theta)

derive the above equation in terms of d(Theta)

you should get:

dx = <derivative of right hand equation>