# Calculus II Trig Sub

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

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

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!

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

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

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

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

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

would d(theta) be 3sec(theta)tan(theta) ?
if you let x = 3tan (theta)

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

is this correct so far?

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

is this correct so far?

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