# Another integration problem

1. Apr 4, 2004

### mathrocks

the integral of tan^3(x^2)sec^4(x^2) dx

I've tried many different methods: getting rid of x^2, splitting up tan^3(x^2) and using the double angle formula, but for some reason it's not working for me. If someone would show me how to set it up it would be greatly appreciated.

2. Apr 4, 2004

### mathrocks

anyone have any ideas?

3. Apr 4, 2004

### Nexus[Free-DC]

Since tan=sin/cos and sec=1/cos express your integral as:

sin^3(x^2)
----------- dx
cos^7(x^2)

Now substitute in u=x^2 and integrate with respect to u (be careful here; don't forget to switch between dx and du!)

Hopefully that will get you started.

PS. Can someone show me how to use Tex here?

4. Apr 4, 2004

### ShawnD

I put that into Maple and it gave me an answer (probably by parts) that is 8 lines long and has an integral at the end. Whatever you do, don't try to integrate it by parts.

I tried to post the output latex but it doesn't work properly.

Last edited: Apr 4, 2004
5. Apr 4, 2004

### mathrocks

But see, if you have u=x^2, you get du=2x dx or 1/2du=x dx which doesn't replace anything in the function.

6. Apr 4, 2004

du = 2x dx

u = x^2, so x = sqrt(u), therefore

du = 2sqrt(u) dx
dx = du/(2sqrt(u))

Which can be substituted in, but that doesn't simplify the integral a whole lot. Mathematica still can't do it.

Just out of curiosity, where do you get these things?

7. Apr 4, 2004

### Zurtex

Last edited: Apr 4, 2004
8. Apr 4, 2004

### mathrocks

It's a take-home quiz i have for my college calc II class

9. Apr 4, 2004

### ShawnD

I don't think that integrator even works. Here is what I input:

$$\int sin^3(x^2) dx$$

Here is what it says the answer is

$$\frac{1}{3}sin^3(x^3)$$

I don't believe that for a second. Here is what Maple says the answer is

$$-1/24\,\sqrt {2}\sqrt {\pi }\sqrt {3}{\it FresnelS} \left( {\frac { \sqrt {2}\sqrt {3}x}{\sqrt {\pi }}} \right) +3/8\,\sqrt {2}\sqrt {\pi }{\it FresnelS} \left( {\frac {\sqrt {2}x}{\sqrt {\pi }}} \right)$$

10. Apr 4, 2004

### Zurtex

Yes it does work it just has a temperamental input system.

When integrating $\sin^3 x^2$ you need to input exactly:

(Sin[x^2])^3

$$\frac{1}{4} \sqrt{\frac{\pi}{6}} \left( 3 \sqrt{3} {\it FresnelS} \left[ \sqrt{ \frac{2}{\pi} } x \right] - {\it FresnelS} \left[\sqrt{ \frac{6}{\pi} } x \right] \right)$$

I only just worked this out my above answer was wrong.

11. Apr 4, 2004

### mathrocks

wow, that's a weird looking answer...

12. Apr 4, 2004

They're seriously giving this problem to you for a Calc II class? What kind of material are you covering?

13. Apr 5, 2004

### harsh

I am not totally sure about this, but cant you use trigonometric integrals to solve this problem. Since its a Calc II class, i think it might be the way to do this. I also think that all Calc II books have a rule when you have tan and sec in a integral, and if i recall correctly, you can sub in tan^2(x^2). I will actually look at this problem in a bit..but you can give this a shot.

14. Apr 5, 2004

The problem is the x^2 inside the trig functions.

15. Apr 5, 2004

### Zurtex

Well I've been trying some wacky substitutions but the best I have so far is:

$$- \int \frac{1-v^2}{v^7} \sqrt{ \arccos (v) } dv$$

16. Apr 5, 2004

### ShawnD

The best I've managed to do is turn everything into tan.
sec^2 (x) = 1 + tan^2(x)

Therefore,

$$\int \tan^3(x^2) \sec^4(x^2) dx= \int \tan^3(x^2) [1 + \tan^2(x^2)]^2 dx$$

$$\int \tan^3(x^2) + 2 \tan^5(x^2) + \tan^7(x^2) dx$$

It doesn't seem to be going anywhere since Maple outputs 8 lines of useless garbage; just like the original form.
Has anybody tried partial fractions yet?

17. Apr 5, 2004

### mathrocks

So far we've covered u-sub, trig sub, integration by parts, and trig intergals. With these types of problems I'm suppose to approach it by breaking up the odd powered trig function and sub in the even powered one. So for example on this one I would have something like:

(sec^2 x^2 -1) tan x^2 sec^4(x^2) dx but again, the x^2 is messing the whole thing up. If it wasn't there I would simply have u=sec^2 (x) and du=tan x and that would make it alot easier.

Last edited: Apr 5, 2004
18. Apr 5, 2004

Heh, I guess the only other thing I can say is are you sure that those x^2's weren't really x2's?

19. Apr 5, 2004

### mathrocks

No, they're x^2...would it make the integration any easier if I have the limits of integration? It's a definite integration problem so the function is from the interval of sqrt(pi/6) and sqrt(pi/4).

20. Apr 5, 2004

### NateTG

$$\int \tan^3(x^2) \sec^4(x^2) dx= \int \tan^3(x^2) \sec^2(x^2) (1+\tan^2(x^2)) \dx$$
$$\int \tan^3(x^2) \sec^2(x^2) dx+ \int \tan^5(x^2) \sec^2(x^2) dx$$
Now it would be nice if we had $$2x$$ in the integrals.
$$\int \tan^3(x^2) \sec^2(x^2) \times 2x \times \frac{1}{2x} dx+ \int \tan^5(x^2) \sec^2(x^2) dx$$
so
$$\frac{\tan^4(x^2)}{8x} + \int \frac{\tan^4(x^2)}{2x^2}dx + \frac{\tan^6(x^2)}{12x} + \int \frac{\tan^6(x^2)}{2x^2}dx$$
$$\frac{\tan^4(x^2)}{8x} + \frac{\tan^6(x^2)}{12x} + \frac{1}{2}\int \frac{\tan^4(x^2)+\tan^6(x^2)}{x^2}dx$$
Now let's try $$x=\sqrt{\arctan \theta}$$,$$dx=\frac{1}{(1+\theta^2)2\sqrt{\arctan \theta}}$$
$$\frac{\tan^4(x^2)}{8x} + \frac{\tan^6(x^2)}{12x} + \frac{1}{2} \int \frac{(1+\theta^2)\theta^4}{2\arctan^{\frac{3}{2}}\theta (1+\theta^2)}d\theta$$
$$\frac{\tan^4(x^2)}{8x} + \frac{\tan^6(x^2)}{12x} + \frac{1}{4} \int \frac{\theta^4}{\arctan^{\frac{3}{2}}\theta} d\theta$$