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Math Help - Integration, Trig substitution

  1. #1
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    Integration, Trig substitution

    Hi I tried these three questions, on trig substitution and I didn't get the answer for all three..I'm sorry I cannot show my whole working, latex is hard for me and I'm struggling trying my best to type the more 'important' parts in it. Hopefully it turns out fine..

    1.) Integrate \frac{1}{x^2\sqrt{16-x^2}}dx
    I had let x be 4\sin\theta, dx = 4\cos\theta d(delta sign)
    Subbing it in and simplifying, I got \frac{1}{16\sin\theta} left to integrate..After integrating I got \frac{-cot\theta}{16}+c and since x = 4\sin\theta, \sin\theta = \frac{x}{4} ..
    Substituting, I nearly got the answer,
    My answer is \frac{\-sqrt{16-x^2}}{4x} + c but it's wrong..

    2.) Integrate \frac{1}{x\(x^2+16)}dx
    I let x be 4\tan\theta, dx = 4\sec^2\theta d(delta sign)
    Subbing in and simplifying, I got to integrating \frac{tan\theta}{16}, used the substitution method to tackle this integration of tan\theta, got \frac{1}{16}\ln|cos\theta|}+c but then I tried to put x in terms of delta, it didn't work out for I get really weird looking answers...!

    3.) Integrate x^2(sqrt{4-x^2}), I had let x be 4\sin\theta, dx = 2\cos\theta d(delta sign)
    Subbing and simplifying, I was left with integrating \16\cos\^2\theta\-\16\cos\^4\theta}dx. Using the double angle expression, I simplified, integrated and got 4\sin\2\theta\-\2\sin\4\theta + c... continued with it but I'm getting nowhere!

    Thank you so much for your time!
    P.s. the trigo functions with values like 2 and 4 with that little ^ above them, sorry for my bad skills, it actually means that it is eg. (cos(x))^2 etc!
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Integration, Trig substitution

    Quote Originally Posted by Tutu View Post
    After integrating I got \frac{-cot\theta}{16}+c and since x = 4\sin\theta, \sin\theta = \frac{x}{4} ..
    Substituting, I nearly got the answer,
    My answer is \frac{\-sqrt{16-x^2}}{4x} + c but it's wrong..
    Almost got it - a minor math error is all that's wrong. The denominator should be 16x, not 4x. If  \sin \theta = \frac x 4, then  \cos \theta = \frac 1 4 \sqrt {16 - x^2} and  \frac {-cot \theta}{16} = - \frac {\frac 1 4 \sqrt {16 - x^2}}{16 \frac x 4} = \frac {- \sqrt {16-x^2}}{16x}

    Quote Originally Posted by Tutu View Post
    2.) ...Subbing in and simplifying, I got to integrating \frac{tan\theta}{16}
    You should have  \frac {cot \theta} {16} here.

    Quote Originally Posted by Tutu View Post
    3.) ...Using the double angle expression, I simplified, integrated and got 4\sin\2\theta\-\2\sin\4\theta + c... continued with it but I'm getting nowhere!
    At this point you should have 4\sin^2\theta-2\sin^4\theta + c. To integrate sin^2x use the identity  \sin^2 \theta = \frac { 1 - \cos(2 \theta)} 2. And to integrate sin^4 \theta treat is as  (sin^2 \theta) ^2

    (sin^2 \theta)^2 = (\frac { 1 - \cos(2 \theta)} 2)^2 = \frac {1 - 2 cos(2 \theta) + \cos^2(2 \theta)} 4 = \frac {1 - 2 cos(2 \theta) + (\frac{1+\cos(4 \theta)}2)} 4 =\frac 3 8 - \frac {cos(2 \theta)} 4 + \frac {\cos(4 \theta)} 8
    Last edited by ebaines; December 12th 2012 at 11:04 AM.
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    Re: Integration, Trig substitution

    Thank you! I got the first one!

    I got \frac{cot}{16} for the second one, and then after integrating and simplifying I got \frac{1}{16}\ln\|\sin\theta|\+c, could you show be how to further simplify?

    For the third one, I could not get what you got.. for 4sin^2(delta)-2sin^4(delta) it seemed that you integrated to get that. But why? Later you performed another integration, so integrate two times?

    Thank you really very much for your time!
    J.
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  4. #4
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    Re: Integration, Trig substitution

    oops, it's (1/16) ln|(sin(delta)| + c, sorry!
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Integration, Trig substitution

    For the second one, you need to replace the  \sin \theta term with the equivalent in terms of  x:

     x = 4 \tan \theta = \frac {4 \sin \theta}{\cos \theta} = \frac {4 \sin \theta} {\sqrt {1 - \sin^2\theta}};

     x^2(1-\sin^2 \theta) = 16 sin^2 \theta

    Rearrange to get:  \sin \theta = \frac x {\sqrt {x^2+16}}

    So the result is

     \int \frac 1 {x (x^2 + 16)} dx = \frac 1 {16} ln(\frac x {\sqrt{x^2 + 16}}) + C

    For the third one - I guess I wasn't very clear. You said you reached a point where you need to integrate  16 \cos^2 \theta - 16 \cos^4 \theta and you ended up with  4 \sin (2 \theta) + 2 \sin (4 \theta) + C. I arrived at a different expression to integrate, and then tried to show how to evaluate that integral. But as I review this I think your approach of integrating  16 \cos^2 \theta - 16 \cos^4 \theta works as well. However, the result of that integral is not what you got - it should actually be:  2 \theta - \frac  1 2 \sin(4 \theta) .
    Last edited by ebaines; December 13th 2012 at 06:55 AM.
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    Re: Integration, Trig substitution

    Thank you so much!

    I got the answer key for the questions, question 2 is right, thank you so much! It's just that it is |x| in the numerator but the denominator was \sqrt{x^2+16}, without the mod. I don't really see why..

    For the third one, I cannot get what you got!
    Here are my steps:

    Simplifying \16\cos^2\theta\-\16\cos^4\theta
    I have \8\cos\2\theta\-\8\cos\4\theta left to integrate
    After integration,

    \frac{8sin\2\theta}{2}-\frac{8sin\4\theta}{4} + c
    = \4sin\2\theta-2sin\4\theta + c
    = \8\sin\theta\cos\theta-4sin\2\theta\cos\2\theta + c
    that's where I'm stuck.
    I think I had performed the integration wrongly..

    Nonetheless, thank you so much!
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  7. #7
    MHF Contributor ebaines's Avatar
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    Re: Integration, Trig substitution

    Quote Originally Posted by Tutu View Post
    Simplifying \16\cos^2\theta - 16\cos^4\theta
    I have 8\cos2\theta-8\cos4\theta left to integrate
    First, please note that some of you math symbols aren't working right, I believe because you are putting a backslash in front of minus signs and numbers where it is not needed.

    I get  16  \cos^2 \theta + 16 \cos^4 \theta = 2  - 2 \cos (4 \theta) , which integrates easily enough to  2 \theta - \frac {\sin (4 \theta)}2
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