Trigonometric substitution integration problem?

Question

from ln(3/4) to ln(4/3)

integrate [(e^t / (1+e^2t)^(3/2)) dt]

I need some serious help on this problem. I got all the simpler problems in this HW, but can't figure this one out. Can anyone help me?

Thanks

Answer 1

Integral ( e^x / [1 + e^(2x)]^(3/2) dx )

To solve this, first note that e^(2x) = (e^x)^2. Change it as such.

Integral ( e^x / [1 + (e^x)^2]^(3/2) dx )

Next, I'm going to move the e^x next to the dx.

Integral ( 1 / [1 + (e^x)^2]^(3/2) e^x dx )

Here's where I'll use substitution.

Let u = e^x. Then
du = e^x dx

Integral ( 1 / [1 + u^2]^(3/2) du

Now, we use trigonometric substitution.

Let u = tan(t).
du = sec^2(t) dt

Integral ( 1 / [1 + tan^2(t)]^(3/2) sec^2(t) dt

Using the identity 1 + tan^2(t) = sec^2(t), we get

Integral ( 1 / [sec^2(t)]^(3/2) sec^2(t) dt

Simplifying, we get

Integral ( 1 / [sec^3(t)] sec^2(t) dt )

Moving the sec^2(t) to the top,

Integral ( sec^2(t)/sec^3(t) dt )

And, we get a reduction.

Integral ( 1/sec(t) dt)

Integral ( cos(t) dt)

Which we can now integrate easily.

sin(t)

To convert this back in terms of u, remember that we let
u = tan(t); this means, by SOHCAHTOA, that
tan(t) = u/1 = opp/adj, so
opp = u
adj = 1
hyp = sqrt(1^2 + u^2) = sqrt(1 + u^2)

Therefore,
sin(t) = opp/hyp = u/sqrt(1 + u^2)

And our answer because

u/sqrt(1 + u^2)

But u = e^t, so our final answer is

[e^t] / sqrt[1 + (e^t)^2] {evaluated from ln(3/4) to ln(4/3) }

= e^[ln(4/3)] / sqrt(1 + [e^[ln(4/3)]]^2) - e^[ln(3/4)] / sqrt(1 + [e^[ln(3/4)]]^2)

= [(4/3) / sqrt(1 + (4/3)^2)] - [(3/4) / sqrt(1 + (3/4)^2)]
= [(4/3) / sqrt(1 + 16/9)] - [(3/4) / sqrt (1 + 9/16)]
= [(4/3) / sqrt(25/9)] - [(3/4) / sqrt (25/16)]
= [(4/3) / (5/3)] - [(3/4) / (5/4)]
= (4/5) - (3/5)
= 1/5

****
As a side note, I didn't have to plug in my bounds for integration as the original. When I reached the step
Let u = e^x, I could have went {when x = ln(4/3), u = 4/3.
when x = ln(3/4), u = 3/4}, and solved the integral with our new bounds with respect to u.

Answer 2

Integrate [(e^t / (1+e^2t)^(3/2)) dt] from ln(3/4) to ln(4/3)

∫{e^t / (1 + e^2t)^(3/2)}dt
Let
tanθ = e^t
sec²θdθ = e^t dt

= ∫{sec²θ / (1 + tan²θ)^(3/2)}dθ
= ∫{sec²θ / (sec²θ)^(3/2)}dθ
= ∫{sec²θ / sec³θ}dθ
= ∫cosθ dθ = sinθ

tanθ = e^t
tan²θ = e^(2t)
1 + tan²θ = sec²θ = 1 + e^(2t)
cos²θ = 1/[1 + e^(2t)]
sin²θ = 1 - cos²θ = 1 - 1/[1 + e^(2t)]
sin²θ = [1 + e^(2t) - 1] / [1 + e^(2t)]
sin²θ = e^(2t) / [1 + e^(2t)]
sinθ = e^(t) / √[1 + e^(2t)]

∫{e^t / (1 + e^2t)^(3/2)}dt
= sinθ
= e^(t) / √[1 + e^(2t)] | [Evaluated from ln(3/4) to ln(4/3)]

= e^(ln 4/3) / √[1 + e^(2ln 4/3))]
- e^(ln 3/4) / √[1 + e^(2ln 3/4))]

= (4/3) / √(1 + 16/9) - (3/4) / √(1 + 9/16)
= (4/3) / √(25/9) - (3/4) / √(25/16)
= (4/3) / (5/3) - (3/4) / (5/4)
= 4/5 - 3/5 = 1/5

Answer 3

(e^t)/[(1+e^(2*t))^(1/2)] You can then evaluate this from the given bounds.

Answer 4

I = x³ / ?(one hundred forty four - x²) permit x = 12 sin ? dx = 12 cos ? d? subsequently, ? x³ dx / ?(one hundred forty four - x²) = ? (12 sin ?)³ (12 cos ? d?) / [?{one hundred forty four - (12 sin ?)²}] = 1728 ? sin³ ? d? You have been superb suited till this factor. enable us to proceed. ? x³ dx / ?(one hundred forty four - x²) = ? (12 sin ?)³ (12 cos ? d?) / [?{one hundred forty four - (12 sin ?)²}] = 1728 ? sin³ ? d? = 1728 ? (sin² ? * sin ?) d? = 1728 ? (a million - cos² ?) * sin ? d? = 1728 ? (sin ? - sin ? cos² ?) d? = 1728 [ - cos ? + (a million/3)cos³ ? ] + c wish this enables! J