Another tan and sec integral issue

[prace]

Here is another problem that might be related to my last post. These dang sec and tan. Anyway, here goes...

$$\int\frac{x^3}\sqrt{{x^2+1}}\dx$$

Oh man.. above is a feeble attempt to write with the LaTeX language. that was my first attempt and a terrible one at that. Haha.. Sorry about that. Anyway, to the question.

I am trying to integrate ∫(x^3)/((x^+1)^(1/2)) dx

To start, I let x = tan(θ) which let's dx = sec^2(θ) dθ
Plugging this back into the equation, we get ∫ (tan^3(θ)/((tan^2(θ)+1)^(1/2)) sex^2(θ) dθ.

Solving the algebra I come to a point that I do not know what to do next.

That is, ∫ tan^3(θ)sec(θ) dθ.

Can anyone offer any advice as to what the next step would be?

Thank you

 

[arildno]

Again, don't bother with trig substitutions!
Let $$u(x)=x^{2}, \frac{dv}{dx}=\frac{x}{\sqrt{x^{2}+1}}$$

 

[prace]

Wait, so if I let u = $$x^2$$, then what does that make the numerator after substituting it back in? Am I missing some fundamental algebra thing here, because I don't see how that works.

 

[HallsofIvy]

The way arildno wrote that, it looks like an "integration by parts" that I don't follow! Let u= x²+ 1 so that du= 2x dx or (1/2)du= xdx.
The numerator, then, is x²(xdx)= (1/2)(u-1)du and the denominator is u^1/2.

Of course, you can do the tan, sec integral: reduce it to sine and cosine:
$$tan^3 \theta sec \theta= \frac{sin^3 \theta}{cos^4 \theta}$$
Write that as
$$\frac{sin^2 \theta}{cos^4 \theta} sin \theta = \frac{1- cos^2 \theta}{cos^4 \theta} sin \theta$$
and let u= cos $\theta$.

(To get the root around the whole thing, use braces: {x^2+ 1}.)

 

[arildno]

Don't follow??
It is perfectly obvious!
We get $v=\sqrt{x^{2}+1}$, thus:
$$\int\frac{x^{3}}{\sqrt{x^{2}+1}}dx=x^{2}\sqrt{x^{2}+1}-\int{2}x\sqrt{x^{2}+1}dx=x^{2}\sqrt{x^{2}+1}-\frac{2}{3}(x^{2}+1)^{\frac{3}{2}}+C$$

 

[prace]

Ok: I tried both approaches here and neither of them seem to be getting me to the correct answer. Or, maybe it is the correct answer and I don't know it. The answer that arildno gave looks correct to me, but I do not follow the algebra. I know, as a calculus student I should be able to follow this no trouble, but it seems a little confusing to me. If you do not want to waste your time explaining it I totally understand.

The solution that HallsofIvy gave me I completely understand, but I don't seem to come out to a correct answer.

After
$$\frac{sin^2 \theta}{cos^4 \theta} sin \theta = \frac{1- cos^2 \theta}{cos^4 \theta} sin \theta$$

and then letting u = cos(θ), I then came out to
$$3sec^3 \theta+sec \theta$$

That solved for θ, so by building a triangle and finding out what θ is I got
$$1+x^2+3(1+x^2)^3\$$

That brings about another fundametal issue that I might have a problem with. For example, if I evaluate an integral and come to an answer like the above, $$3sec^3 \theta+sec \theta$$, for the the $$3sec^3 \theta$$ part, can I find sec(θ) and then just cube it? If I can, then maybe I got the right answer, but if you can't, then I suppose I am way off. My guess is that you can't.

 

[VietDao29]

After
$$\frac{sin^2 \theta}{cos^4 \theta} sin \theta = \frac{1- cos^2 \theta}{cos^4 \theta} sin \theta$$

and then letting u = cos(θ), I then came out to
$$3sec^3 \theta+sec \theta$$

This line is incorrect.
You should note that:
$$\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \ \alpha \neq -1$$
Now:
$$\int \frac{1 - \cos ^ 2 \theta}{\cos ^ 4 \theta} \sin \theta d\theta = - \int \frac{1 - \cos ^ 2 \theta}{\cos ^ 4 \theta} d(\cos \theta) = - \int \frac{d(\cos \theta)}{\cos ^ 4 \theta} + \int \frac{d(\cos \theta)}{\cos ^ 2 \theta}$$
$$= - \int \cos ^ {-4} \theta d(\cos \theta) + \int \cos ^ {-2} \theta d(\cos \theta) = \frac{1}{3 \cos ^ 3 \theta} - \frac{1}{\cos \theta} + C$$
Yes, you can convert $$\theta$$ back to x like this:
Since you let:
$$x = \tan \theta \Rightarrow x ^ 2 + 1 = \tan ^ 2 \theta + 1 = \sec ^ 2 \theta \Rightarrow \sec \theta = \sqrt{x ^ 2 + 1}$$
So:
$$\sec \theta = \sqrt{x ^ 2 + 1}$$. Can you change $$\theta$$ back to x now? :)
--------------------
Here's another method (the u-substitution one)
Now we must find $$\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}}$$, we have:
$$\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}} = \frac{1}{2} \int \frac{x ^ 2 d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}}$$ (make the substitution u = x² + 1)
$$= \frac{1}{2} \int \frac{(x ^ 2 + 1 - 1) d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}} = \frac{1}{2} \int \sqrt{x ^ 2 + 1} d(x ^ 2 + 1) - \frac{1}{2} \int \frac{d(x ^ 2 + 1)}{\sqrt{x ^ 2 + 1}}$$.
Can you go from here? :)
--------------------
About arildno's method:
Integration by Parts is:
$$\int u dv = uv - \int v du$$
$$\int \frac{x ^ 3 dx}{\sqrt{x ^ 2 + 1}}$$
We then let u = x², and $$dv = \frac{x dx}{\sqrt{x ^ 2 + 1}}$$
u = x² => du = 2x dx
$$dv = \frac{x dx}{\sqrt{x ^ 2 + 1}} \Rightarrow v = \sqrt{x ^ 2 + 1}$$
Substitute all those to our formula, and we are done. Can you get this? :)