# More Trig Subsitution

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.

## Homework Statement

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

## Homework Equations

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

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

## 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:

Dick
Homework Helper
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.

VietDao29
Homework Helper
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. :)

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

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:
Gib Z
Homework Helper
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 $$\sqrt{u+1}$$..

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.

Gib Z
Homework Helper
$$\int \frac{1}{x} dx = \ln x$$

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.

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