#### ChaoticLlama

##### Junior Member

- Joined
- Dec 11, 2004

- Messages
- 199

Reduction:

∫e^(5*x) * cos(3x)^2

Trig Substitution:

∫x²dx / √(x² - 1)

thanks for your time

- Thread starter ChaoticLlama
- Start date

- Joined
- Dec 11, 2004

- Messages
- 199

Reduction:

∫e^(5*x) * cos(3x)^2

Trig Substitution:

∫x²dx / √(x² - 1)

thanks for your time

I hope you know the trig identitiesTrig Substitution:

∫x²dx / √(x² - 1)

sec^2 t = 1 + tan^2 t

cot^2(x) + 1 = csc^2(x)

Substitute in place of X .

Come back with a re-post if you get stuck.

- Joined
- Dec 11, 2004

- Messages
- 199

let x = tan[θ]

dx = sec[θ]^2dθ

∫(tan[θ]^2 * sec[θ]^2)dθ / tan[θ]

∫(tan[θ] * sec[θ]^2)dθ

Once I get here i'm not sure when to integrate the trig functions.

You did okay up to a point . . .

The radical becomes: .tan θ∫x²dx / √(x² - 1)

let x = tan θ

dx = sec<sup>2</sup> θ dθ

. . . . . . . . . . . . . . . . . . . .sec<sup>2</sup>θ

The integral becomes: . --------- (sec θ tan θ dθ) . = . sec<sup>3</sup>θ dθ

. . . . . . . . . . . . . . . . . . . .tan θ

The dreaded "secant-cubed" requires a repeated by-parts solution.

. . . The answer is: . (1/2) [sec θ tan θ + ln|sec θ + tan θ|] + C

And I'll let