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Math Help - Trig substitution

  1. #16
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    Okay, I have that so far,

    I have: 64 integral from (pi/4) to (pi/3) (1+2cos(2theta)+cos(2theta)^2) dtheta. Then I split those up into the integrals (still multiplying everything by 64) of 1, 2cos(2theta) and cos(2theta)^2. I have a lot of work shown, but I ended up getting the answer 64(3pi/32+sin(pi/6)+1/8sin(pi/3). That answer isn't right and I simplified it to the best of my ability and I don't know where I went wrong! But for the integral of 1, I got (pi/3-pi/4) and for the integral of 2cos(2theta) I did u substitution and got sin(2theta) and used the limits, and then for cos(2theta)^2 I used the half angle formula and got (1+cos(4theta) over 2, since its double that of a single half angle formula. Using the formula, I got (1/8) integral of 1 + integral sin(4theta). SO that was for the last integral when I first split everything up.
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  2. #17
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    Quote Originally Posted by elocin View Post
    I have: 64 integral from (pi/4) to (pi/3) (1+2cos(2theta)+cos(2theta)^2) dtheta. Then I split those up into the integrals (still multiplying everything by 64) of 1, 2cos(2theta) and cos(2theta)^2. I have a lot of work shown, but I ended up getting the answer 64(3pi/32+sin(pi/6)+1/8sin(pi/3). That answer isn't right and I simplified it to the best of my ability and I don't know where I went wrong! But for the integral of 1, I got (pi/3-pi/4) and for the integral of 2cos(2theta) I did u substitution and got sin(2theta) and used the limits, and then for cos(2theta)^2 I used the half angle formula and got (1+cos(4theta) over 2, since its double that of a single half angle formula. Using the formula, I got (1/8) integral of 1 + integral sin(4theta). SO that was for the last integral when I first split everything up.
    Its hard to read everything, because is not in latex, but where did that 64 come from

    We alread had

    \frac{1}{256} before we even simplified \cos^4(\theta)
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  3. #18
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    Well (1) / (1/4)^4 equals 256, and I got 64 because after multiplying the 2 half angle formulas of cosine, I took out that (1/4) and multiplied 256 by it. After multiplying (1+cos(2theta)/2 by itself, you get (1+2cos(2theta)+cos(2theta)^2 over 4. I took out that 4 (in terms of 1/4) and multiplied 256 by it. I don't know if that makes sense. I don't know what latex is! I'm sorry!
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  4. #19
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elocin View Post
    Well (1) / (1/4)^4 equals 256, and I got 64 because after multiplying the 2 half angle formulas of cosine, I took out that (1/4) and multiplied 256 by it. After multiplying (1+cos(2theta)/2 by itself, you get (1+2cos(2theta)+cos(2theta)^2 over 4. I took out that 4 (in terms of 1/4) and multiplied 256 by it. I don't know if that makes sense. I don't know what latex is! I'm sorry!
    Ok, I see what you did I misinterpreted your last post. So are you sure that you are wrong?
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  5. #20
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    Yeah, well I put the answer: 64(3pi/32+sin(pi/6)+1/8sin(pi/3)), if you can read that, I'm not sure how to use this symbols thing for the math forum!! My computer says it isn't the right answer! Does that look right to you?
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  6. #21
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elocin View Post
    Yeah, well I put the answer: 64(3pi/32+sin(pi/6)+1/8sin(pi/3)), if you can read that, I'm not sure how to use this symbols thing for the math forum!! My computer says it isn't the right answer! Does that look right to you?
    Ill just do it

    64\int\cos^4(\theta)d\theta=32\int\bigg[1+2\cos(\theta)+\cos^2(\theta)\bigg]d\theta= 32\int\bigg[1+2\cos(\theta)+\frac{1}{2}+\frac{1}{2}\cos(4\thet  a)\bigg]d\theta

    Using basic integration techniques I get

    34\bigg[\frac{3\theta}{2}+2\sin(\theta)+\frac{1}{8}\sin(4\  theta)\bigg]


    and math \theta=arcsec\left(4x\right)

    \theta\left(\sqrt{\frac{2}{4}}\right)=\cdots

    Here is where I think our problem is you have the bounds as

    \frac{\sqrt{2}}{4}\text{ and }\frac{1}{2}

    The left one cannot be right, is it?
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  7. #22
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    Well (radical 2 all over 4) and (1/2) are the bounds from the original problem, and I used cos(30degrees) and cos(45degrees) to convert to theta, giving me (pi/3) and (pi/4). So for (1/2), I set that equal to (1/4)sectheta, then divided (1/2) by (1/4) and got 2. Then I set 2=(1)/(costheta) since that is the same thing as sectheta, and then got costheta=1/2. Then I got the angle 60 I think and that got me pi over 3. Thats how I did the other one too! Maybe that was wrong?
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  8. #23
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    Yeah it's only the radical 2 and thats over 4. It isn't the whole radical of 2/4.
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  9. #24
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elocin View Post
    Well (radical 2 all over 4) and (1/2) are the bounds from the original problem, and I used cos(30degrees) and cos(45degrees) to convert to theta, giving me (pi/3) and (pi/4). So for (1/2), I set that equal to (1/4)sectheta, then divided (1/2) by (1/4) and got 2. Then I set 2=(1)/(costheta) since that is the same thing as sectheta, and then got costheta=1/2. Then I got the angle 60 I think and that got me pi over 3. Thats how I did the other one too! Maybe that was wrong?
    No no you are right, I forgot to multiply by the four, My bad compoletely sorry. Let me look back through your work and get back to you, ok ,because when I evaluate from what I got from your work its wrong. So either I made a copying mistake or you made a mistake in your orginal computations, I'll report back as soon as I am able
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  10. #25
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    Oh thank you!! I really appreciate it!!
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  11. #26
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    \int\frac{dx}{x^5\sqrt{16x^2-1}}

    let 4x=\sec(\theta)\Rightarrow\frac{\sec(\theta)\tan(\  theta)}{4}=dx}

    So we have

    \int\frac{\frac{1}{4}\sec(\theta)\tan(\theta)}{\se  c^5(\theta)\frac{1}{4^5}\tan(\theta)}d\theta=256\i  nt\cos^4(\theta)d\theta

    Now if we expand we get

    64\int\bigg[\frac{3}{2}+2\cos(2\theta)+\frac{1}{2}\cos(4\theta  )\bigg]d\theta=64\bigg[\frac{3\theta}{2}+\sin(2\theta)+\frac{1}{8}\sin(4\  theta)\bigg]

    Now we must evaluate this from

    \theta\left(\frac{\sqrt{2}}{4}\cdot{4}\right)=\fra  c{\pi}{4}

    To

    \theta\left(\frac{1}{2}\cdot{4}\right)=\frac{\pi}{  3}

    So we would have

    64\bigg[\frac{3\theta}{2}+\sin(2\theta)+\frac{1}{8}\sin(4\  theta)\bigg]\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{3}}=28\sqrt{3}+  8\pi-64

    Which is the right answer? Where did you mess up?
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  12. #27
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    Yep, you're right!! To tell you the truth, I think I messed up the integral of 1!! That's embarrassing. I think that's where I went wrong, because everything else looks the same, except for the 3theta over 2. I had 3theta over 32, which I got from adding (pi/12) to (pi/96). Yeah I did that wrong. Thank you SO much, though for taking the time!!! Now I have to study this so I get it right next time!!
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  13. #28
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elocin View Post
    Yep, you're right!! To tell you the truth, I think I messed up the integral of 1!! That's embarrassing. I think that's where I went wrong, because everything else looks the same, except for the 3theta over 2. I had 3theta over 32, which I got from adding (pi/12) to (pi/96). Yeah I did that wrong. Thank you SO much, though for taking the time!!! Now I have to study this so I get it right next time!!
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