Can Integration by Parts Solve This Integral Problem?

[Knissp]

Homework Statement

\int_0^1 (6t^2 (1+9t^2)^{1/2} dt)

Homework Equations

\int u dv = u v - \int v du

The Attempt at a Solution

\int_0^1 (6t^2 (1+9t^2)^{1/2} dt)
=6 * \int (t^2 (1+9t^2)^{1/2} dt)
= 6 * \int (t * t (1+9t^2)^{1/2} dt)

Let u = t; let dv = t (1+9t^2)^{1/2} dt;
then du = dt; and v = \int t (1+9t^2)^{1/2} dt

(using w-substitution:
w = 1+9t^2,
dw = 18t dt;
dw/18=dt;
\int t (1+9t^2)^{1/2} dt
=1/18 \int w^{1/2} dw = 1/18 * 2/3 w^{3/2} = w^{3/2} / 27 = (1+9t^2)^{3/2}/27

v = [(1+9t^2)^{3/2}/27

\int(u dv) = u v - \int (v du)
= t * (1+9t^2)^{3/2}/27 - \int ((1+9t^2)^{3/2}/27 dt)

now i need help integrating (1+9t^2)^{3/2}/27.

 

[tiny-tim]

Hi Knissp!

The Attempt at a Solution

…

oooh, that's horrible!

Hint: start again, and make the obvious substitution at the beginning.

 

[rocomath]

Can you use trig-sub or are you required to solve it by int. by parts?

 

[Knissp]

I can use any method, I just thought by parts would be easiest. What trig subst. would I use?

and Tim: would the obvious substitution be u=t^2 and dv = sqrt(1+9t^2) dt? Then I have to integrate sqrt(1+9t^2) dt...

 

[rocomath]

I can use any method, I just thought by parts would be easiest. What trig subst. would I use?

and Tim: would the obvious substitution be u=t^2 and dv = sqrt(1+9t^2) dt? Then I have to integrate sqrt(1+9t^2) dt...

t=3\tan\theta

 

[Knissp]

t=3\tan\theta
dt = 3\sec^2\theta d\theta

\int (6t^2 (1+9t^2)^{1/2} dt)
= \int (6(3\tan\theta )^2 (1+9(3\tan\theta)^2)^{1/2} 3\sec^2\theta d\theta)
=\int (54 \tan^2\theta (1+81 \tan^2\theta)^{1/2} 3\sec^2\theta d\theta)
=162 \int (\tan^2\theta (1+81 \tan^2\theta)^{1/2} \sec^2\theta d\theta)

Should I keep going from here?

 

[rocomath]

OMG! I'm so sorry ... lol

First, you have to simplify it ... to find what your "a" should be.

 

[gabbagabbahey]

hmm... I think rocomath meant to say 3t=tan(\theta) ;0)

 

[rocomath]

1+9t^2

Dividing by 9, \sqrt{9\left(\frac 1 9+t^2\right)}=3\sqrt{\frac 1 9+t^2}

t=\frac 1 3\tan\theta

 

[Knissp]

Now I have
\int 2/9 \tan^2\theta \sec^3\theta d\theta

which is

\int 2/9 \frac{\sin^2\theta}{\cos^5\theta} d\theta

 

[gabbagabbahey]

Try Integration by parts now (don't forget that your limits of integration are tan^{-1}(0) \rightarrow tan^{-1}(3) now)

 

[tiny-tim]

Now I have
\int 2/9 \tan^2\theta \sec^2\theta d\theta …

Hi Knissp!

(have a theta: θ and a squared: ² and a cubed: ³ and an integral: ∫ )

Isn't it tan²θsec³θ?

 

[Knissp]

I have tried several ways none of which worked, too much to post it all. Just a little more help needed: would it be best to use the identity 1+tan²θ = sec²θ before integrating by parts? edit: just to add, in case anyone wanted to know where this problem is from,

Let C be the curve represented by the equations
x=2t, y=3t² (0≤t≤1) dx/dt=2, dy/dt=6t
Evaluate the line integral along C: ∫(x-y)ds
ds = √((dx/dt)^2+(dy/dt)^2) dt = √(4+36t^2) = 2√(1+9t^2) dt
∫(x-y)ds = ∫((2t-3t^2)*2√(1+9t^2) dt = ∫((4t√(1+9t^2) dt - ∫((6t^2)√(1+9t^2) dt
I was able to solve the ∫((4t√(1+9t^2) dt part using u-substitution, but I got stuck on ∫((6t^2)√(1+9t^2) dt which is why I originally asked for help. Question: Is my way of attempting this problem the most efficient, or is there a better way so I can avoid this integral which involves trig-substitution entirely?

 

[gabbagabbahey]

Use the Identity sin^2(\theta)=1-cos^2(\theta) to break the integral into two parts (sec^3 and sec^5). Then use integration by parts with u=sec(\theta) and dv=sec^2(\theta)d\theta...you will have to use by parts twice for the sec^5 term...you'll also need to look up the integral of sec(\theta)d\theta

 

[Knissp]

tan²θsec³θ = (sec²θ-1)sec³θ = \sec^5\theta - \sec^3\theta
Looking at the sec^3 part, let u = secθ, dv=sec²θ dθ; du=secθtanθ, v=tanθ.
uv-∫vdu = secθtanθ - ∫tan²θsecθ dθ
Is that right so far? because then I have to make another u-sub...

 

[gabbagabbahey]

looks fine so far; now there is a trick you need to use...
-∫tan²θsecθ dθ =∫(1-sec²θ)secθ dθ =∫secθ dθ-∫secθ^3 dθ
=> 2 ∫secθ^3 dθ=secθtanθ+∫secθ dθ

 

[Knissp]

Aaahhh I get it now. Sorry that so long. Thank you all for your help!

 

[tiny-tim]

hyperbolic trig substitution

Hint: start again, and make the obvious substitution at the beginning.

t=\frac 1 3\tan\theta

hmm … when you have a square, there's always the choice of a trig substitution, or a hyperbolic trig substitution …

in this case, either t = (1/3)tanθ, or t = (1/3)sinhu …

using the tan worked, but it was rather complicated.

Try it again, using the sinh instead!