# Integration (by parts)

1. Oct 12, 2008

### Knissp

1. The problem statement, all variables and given/known data
$$\int_0^1 (6t^2 (1+9t^2)^{1/2} dt)$$

2. Relevant equations
$$\int u dv = u v - \int v du$$

3. 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 $$(1+9t^2)^{3/2}/27$$.

2. Oct 12, 2008

### tiny-tim

Hi Knissp!
oooh, that's horrible!

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

3. Oct 12, 2008

### rocomath

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

4. Oct 12, 2008

### 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...

5. Oct 12, 2008

### rocomath

$$t=3\tan\theta$$

6. Oct 12, 2008

### 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?

7. Oct 12, 2008

### rocomath

OMG!!! I'm so sorry ... lol

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

8. Oct 12, 2008

### gabbagabbahey

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

9. Oct 12, 2008

### rocomath

$$1+9t^2$$

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

$$t=\frac 1 3\tan\theta$$

10. Oct 12, 2008

### Knissp

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

which is

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

11. Oct 12, 2008

### gabbagabbahey

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

12. Oct 12, 2008

### tiny-tim

Hi Knissp!

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

Isn't it tan²θsec³θ?

13. Oct 12, 2008

### 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?

Last edited: Oct 12, 2008
14. Oct 12, 2008

### 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 $sec(\theta)d\theta$

15. Oct 12, 2008

### 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...

16. Oct 12, 2008

### 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

17. Oct 12, 2008

### Knissp

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

18. Oct 13, 2008

### tiny-tim

hyperbolic trig substitution

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!