# More Trig Subsitution

1. Mar 14, 2007

### teneleven

Aye gents, I've been working on a problem that doesn't have a solution at the back of the book and would be grateful if you'd compare my answer with yours.

1. The problem statement, all variables and given/known data

$$\int\frac{t^5}{\sqrt{t^2 + 2}}\ dt$$

2. Relevant equations

$$t = \sqrt{2}\tan\theta$$

$$dt = \sqrt{2}\sec^2\theta\ d\theta$$

3. The attempt at a solution

$$\int\frac{(\sqrt{2}\tan\theta)^5}{\sqrt{(\sqrt{2}\tan\theta)^2 + 2}}\ \sqrt{2}\sec^2\theta\ d\theta$$

$$\int\frac{8\tan^5\theta\sec^2\theta}{\sqrt{2\tan^2\theta + 2}\ d\theta}$$

$$8\int\frac{\tan^5\theta\sec^2\theta}{\sqrt{2}\sec\theta}\ d\theta$$

$$\frac{8}{\sqrt{2}}\int\tan^5\theta\sec\theta\ d\theta$$

$$\frac{8}{\sqrt{2}}\int\frac{\sin^5\theta}{\cos^6\theta}\ d\theta$$

$$\frac{8}{\sqrt{2}}\int\frac{[(1 - \cos^2\theta)(1 - \cos^2\theta)]}{\cos^6\theta}\ \sin\theta\ d\theta$$

$$\frac{8}{\sqrt{2}}\int[\frac{1}{\cos^6\theta} - \frac{2}{\cos^4\theta} + \frac{1}{\cos^2\theta}]\ \sin\theta\ d\theta$$

$$\frac{8}{\sqrt{2}}\ [\frac{-1}{7}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^7) + \frac{2}{5}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^5) - \frac{1}{3}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^3)]\ +\ C$$

Thanks.

Last edited: Mar 14, 2007
2. Mar 14, 2007

### Dick

Where the heck are all of the logs coming from??? Besides, you don't even need to do a trig substitution for this one. Just put u=t^2+2.

3. Mar 14, 2007

### VietDao29

As Dick has pointed out, you are all correct from the start, except the last part, why should ln appear in that expression?
It should be continued by letting u = cos x like this:

$$...= - \frac{8}{\sqrt{2}} \int \left( \frac{1}{\cos ^ 6 \theta} - \frac{2}{\cos ^ 4 \theta} + \frac{1}{\cos ^ 2 \theta} \right) d(\cos (\theta)) = \frac{8}{\sqrt{2}} \left( \frac{1}{5 \cos ^ 5 \theta} - \frac{2}{3 \cos ^ 3 \theta} + \frac{1}{\cos \theta} \right) + C$$

Now, change everything back to t.

$$t = \sqrt{2} \tan \theta , \quad \theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right]$$

$$\Rightarrow \tan \theta = \frac{t}{\sqrt{2}}$$

Note that $$\theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right] \Rightarrow \cos \theta \geq 0$$, so:

$$\cos \theta = \sqrt{\frac{1}{1 + \tan ^ 2 \theta}} = \sqrt{\frac{2}{2 + t ^ 2}}$$

Just plug all that in and see what you get. :)

-----------------

Or, even faster, you can do it like this:
$$\int \frac{t ^ 5}{\sqrt{t ^ 2 + 2}} dt = \frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)$$

$$= \int t ^ 4 d(\sqrt{t ^ 2 + 2}) = \int \left( (\sqrt{t ^ 2 + 2}) ^ 2 - 2 \right) ^ 2 d(\sqrt{t ^ 2 + 2}) = ...$$
Can you go from here? :)

Last edited: Mar 14, 2007
4. Mar 14, 2007

### HallsofIvy

Staff Emeritus
That looks like the hard way to me. Rewrite the integral as
$\int \frac{t^5}{\sqrt{t^2+ 1}dt= n\int \frac{(t^2)^2}{\sqrt{t^2+1} (dt)$so, with u= t2+1[/itex]] we have
$$\int \frac{(u-1)^2}{u^{a+1)}du$$

5. Mar 15, 2007

### teneleven

What's wrong with having natural logs in the solution? How does one know to avoid ln when approaching this problem?

I tried Dick's substitution of $$u = t^2 + 2$$ and didn't get very far.

$$u = t^2 + 2$$

$$du = 2t\ dt$$

$$\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t$$

Integration by parts after this step looks like something I want to avoid. Maybe I'm a bit unclear of how to use Dick's substitution?

Thank you for the ideas, VietDao, but I'm thrown off by the suggestion...

$$\frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)$$

Shouldn't the $$d(t^2 + 2)$$ read $$d(2t)$$ ?

HallsofIvy, I'm having trouble understanding how the following is legal using a substitution of $$u = t^2 + 1$$:

$$\int\frac{(u-1)^2}{u^(^a^+^1^)}du$$

Thanks.

Last edited: Mar 15, 2007
6. Mar 15, 2007

### Gib Z

He never said anything was wrong with Natural logs in a solution, maybe he was pointing out that they just werent correct at all.

Dick's suggestion should go more like this.
You have the right set up, but forgot the dt: $$\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t dt$$

Now We make the substitution, u=t^2 + 2, so the denominator is sqrt of u!,

the 2t dt, as you correctly wrote, is du.

t^4 is (u-2)^2.

So you get the pleasure of doing
$$\frac{1}{2}\int \frac{(u-2)^2}{\sqrt{u}} du$$ instead :D

VietDao's suggestion is the same as dicks, but we just wrote the t^2+2 as u, didnt we :D

I don't understand halls method either..assuming he left the 1/2 outside the integral out, I understand his numerator and differential but not the denominator..I thought that should be $$\sqrt{u+1}$$..

7. Mar 15, 2007

### teneleven

Why is ln wrong? If I have the following problem...

$$\frac{1}{2}\int\frac{\sin\theta}{cos^2\theta}\ d\theta$$

my first inclination is to solve it as follows:

$$\frac{1}{2}\ln(cos^2\theta) + C$$

Fill me in.

8. Mar 15, 2007

### Gib Z

$$\int \frac{1}{x} dx = \ln x$$

But $$\int \frac{1}{x^2} =-\frac{1}{x}$$

Have you seen the power rule for differentiation? Use the reverse power rule for integration, except when the power is -1, that is the only case when the reverse power rule does not work and you have to use the Natural Log function.

9. Mar 15, 2007

### Gib Z

In fact, to see why you are so obviously wrong, try differentiating your result, you won't get your integrand, thats for sure.