Calc. II, Integral of tanx(secx)^3/2?

Question

Any ideas on how I'd solve this?

tanx(secx)^3/2

Thank you very much.

Answer 1

(secx)^(3/2) i can write secx*(secx)^(1/2)
so int= mean integral
int tanx(secx)^(3/2)dx=
int tanxsecx(secx)^(1/2)dx

let u=secx
du=tanxsecxdx
so
int tanxsecx(secx)^(1/2)dx =
int u^(1/2)du
answer is (2/3)*u^(3/2)
but u=secx so put it back
answer is
(2/3)*(secx)^(3/2) +C
good luck

Answer 2

Find the integral of tanx(secx)^(3/2).

First rearrange the terms to make it easier to integrate.

tanx(secx)^(3/2) = [tanx(secx)]√secx

Now we can integrate.

∫tanx(secx)^(3/2)dx
= ∫[tanx(secx)](√secx)dx
= (2/3)(secx)^(3/2) + C

Answer 3

INT { tanx(secx)^3/2 dx }

= INT { sec(x)^1/2 * sec(x) * tan(x) dx }

let y =sec(x)

=> dy = sec(x) * tan(x) * dx

Therefore, our integral becomes:

INT { y^1/2 * dy }

= { y^(3/2) } / (3/2) + C

= 2*sec(x)^(3/2) / 3 + C

Answer 4

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Answer 5

Integral ( tan(x) [sec(x)]^(3/2) dx )

Your first step is to decompose [sec(x)]^(3/2) into
[sec(x)]^(1/2) * sec(x)

Integral ( tan(x) [sec(x)]^(1/2) sec(x) dx )

Now, group the sec(x) and tan(x) together, next to the dx.

Integral ( [sec(x)]^(1/2) sec(x) tan(x) dx)

Here's where you use substitution.

Let u = sec(x). Then
du = sec(x) tan(x) dx {which is the tail end of our integral}. so our integral becomes

Integral ( u^(1/2) du)

Which is now easily integrable.

(2/3)u^(3/2) + C.

Substituting u = sec(x) back, we get

(2/3) [sec(x)]^(3/2) + C