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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.

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- #1

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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

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anyone have any ideas?

- #3

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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

ShawnD

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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.Nexus[Free-DC] said:Since tan=sin/cos and sec=1/cos express your integral as:

sin^3(x^2)

----------- dx

cos^7(x^2)

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

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- #5

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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.Nexus[Free-DC] said:

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?

- #6

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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?

cookiemonster

- #7

Zurtex

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I have no idea how you do this, I would have chosen to go for by-parts.

http://integrals.wolfram.com/index.en.cgi gives the answer as being:

Edit: Sorry input it wrong

http://integrals.wolfram.com/index.en.cgi gives the answer as being:

Edit: Sorry input it wrong

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- #8

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It's a take-home quiz i have for my college calc II classcookiemonster said:

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?

cookiemonster

- #9

ShawnD

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I don't think that integrator even works. Here is what I input:Zurtex said:I have no idea how you do this, I would have chosen to go for by-parts.

http://integrals.wolfram.com/index.en.cgi gives the answer as being:

[tex]\frac{1}{5} \sec^4 \left( \tan^3 x^5 \right)[/tex]

[tex]\int sin^3(x^2) dx[/tex]

Here is what it says the answer is

[tex]\frac{1}{3}sin^3(x^3)[/tex]

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

[tex]-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)[/tex]

- #10

Zurtex

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Yes it does work it just has a temperamental input system.ShawnD said:I don't think that integrator even works. Here is what I input:

[tex]\int sin^3(x^2) dx[/tex]

Here is what it says the answer is

[tex]\frac{1}{3}sin^3(x^3)[/tex]

When integrating [itex] \sin^3 x^2[/itex] you need to input exactly:

(Sin[x^2])^3

And gives the answer:

[tex]\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) [/tex]

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

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wow, that's a weird looking answer...Zurtex said:Yes it does work it just has a temperamental input system.

When integrating [itex] \sin^3 x^2[/itex] you need to input exactly:

(Sin[x^2])^3

And gives the answer:

[tex]\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) [/tex]

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

- #12

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cookiemonster

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- #14

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

cookiemonster

cookiemonster

- #15

Zurtex

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[tex] - \int \frac{1-v^2}{v^7} \sqrt{ \arccos (v) } dv [/tex]

- #16

ShawnD

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sec^2 (x) = 1 + tan^2(x)

Therefore,

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

[tex]\int \tan^3(x^2) + 2 \tan^5(x^2) + \tan^7(x^2) dx[/tex]

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

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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:cookiemonster said:

cookiemonster

(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.

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cookiemonster

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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).cookiemonster said:

cookiemonster

- #20

NateTG

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How about:ShawnD said:[tex]\int \tan^3(x^2) \sec^4(x^2) dx= \int \tan^3(x^2) [1 + \tan^2(x^2)]^2 dx[/tex]

[tex]\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[/tex]

[tex]\int \tan^3(x^2) \sec^2(x^2) dx+ \int \tan^5(x^2) \sec^2(x^2) dx[/tex]

Now it would be nice if we had [tex]2x[/tex] in the integrals.

[tex]\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[/tex]

so

[tex] \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[/tex]

[tex] \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 [/tex]

Now let's try [tex]x=\sqrt{\arctan \theta}[/tex],[tex]dx=\frac{1}{(1+\theta^2)2\sqrt{\arctan \theta}}[/tex]

[tex] \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[/tex]

[tex] \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[/tex]

- #21

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NateTG said:How about:

[tex]\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[/tex]

[tex]\int \tan^3(x^2) \sec^2(x^2) dx+ \int \tan^5(x^2) \sec^2(x^2) dx[/tex]

Now it would be nice if we had [tex]2x[/tex] in the integrals.

[tex]\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[/tex]

so

[tex] \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[/tex]

[tex] \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 [/tex]

Now let's try [tex]x=\sqrt{\arctan \theta}[/tex],[tex]dx=\frac{1}{(1+\theta^2)2\sqrt{\arctan \theta}}[/tex]

[tex] \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[/tex]

[tex] \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[/tex]

Thanks for all the help guys, but my professor made a mistake. There's suppose to be an 'x' in front of that function...makes it sooo much easier!

- #22

Zurtex

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- #23

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visit:Nexus[Free-DC] said:PS. Can someone show me how to use Tex here?

https://www.physicsforums.com/misc/howtolatex.pdf

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!!

That's one heck of a mistake!

cookiemonster

That's one heck of a mistake!

cookiemonster

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