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Math Help - [SOLVED] trig subs

  1. #1
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    [SOLVED] trig subs

    <br /> <br />
\int\sqrt{25-t^2}\,dt<br />


    a = 5 u = t so....

    t = 5 sin \theta

    sqrt(25-(5 sin \theta)^2 )

    sqrt(25 cos^2 \theta)

    25 cos \theta

    25 sin \theta

    25 sin(t/5) is what I get after substituting in for t. I know this isn't right but could someone explain why?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    <br /> <br />
\int\sqrt{25-t^2}\,dt<br />


    a = 5 u = t so....

    t = 5 sin \theta

    sqrt(25-(5 sin \theta)^2 )

    sqrt(25 cos^2 \theta)

    25 cos \theta

    25 sin \theta

    25 sin(t/5) is what I get after substituting in for t. I know this isn't right but could someone explain why?
    \sqrt{25} \ne 25
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    Quote Originally Posted by Jhevon View Post
    \sqrt{25} \ne 25

    AHHHHHH ok so lets keep working

    5 cos \theta

    so

    5 sin \theta

    I still get

    5 sin(t/5)

    which is definately nowhere near the answer in the back of my text.

    they have....
    25/2 sin^-1(t/5) + (t * sqrt(25-t^2)/2 + C
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    AHHHHHH ok so lets keep working

    5 cos \theta

    so

    5 sin \theta

    I still get

    5 sin(t/5)

    which is definately nowhere near the answer in the back of my text.

    they have....
    25/2 sin^-1(t/5) + (t * sqrt(25-t^2)/2 + C
    again, you back substituted wrong. \theta \ne \frac t5. it is \sin \theta = \frac t5.

    secondly, where is your d \theta ?!


    lets start over....

    \int \sqrt{25 - t^2}~dt

    let t = 5 \sin \theta

    \Rightarrow \color{red} dt = 5 \cos \theta ~d \theta

    so our integral becomes:

    \int \sqrt{25 - 25 \sin^2 \theta} \cdot {\color{red}5 \cos \theta} ~d \theta = 25 \int \cos^2 \theta ~d \theta

    now finish up
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  5. #5
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    scratch that I think you edited.


    Ok I think I got lost in that last statement because shouldnt this

    sqrt(25 cos^2 theta) 5 cos theta dtheta

    be

    5 cos theta 5 cos theta dtheta

    =

    25 int(cos^2 theta) dtheta?
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    Quote Originally Posted by Jhevon View Post
    again, you back substituted wrong. \theta \ne \frac t5. it is \sin \theta = \frac t5.

    secondly, where is your d \theta ?!


    lets start over....

    \int \sqrt{25 - t^2}~dt

    let t = 5 \sin \theta

    \Rightarrow \color{red} dt = 5 \cos \theta ~d \theta

    so our integral becomes:

    \int \sqrt{25 - 25 \sin^2 \theta} \cdot {\color{red}5 \cos \theta} ~d \theta = 25 \int \cos^2 \theta ~d \theta

    now finish up
    ok

    25/2 sin^-1(t/5) (1 + cos(2theta) ) / 2

    trig integrals always give me a hard time. Did I substitute in correctly for this? If so what would I do with that 1 + cos(2theta) / 2
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    ok

    25/2 sin^-1(t/5) (1 + cos(2theta) ) / 2

    trig integrals always give me a hard time. Did I substitute in correctly for this? If so what would I do with that 1 + cos(2theta) / 2
    no you didn't. you didn't even integrate correctly

    25 \int \cos^2 \theta ~d \theta = 25 \int \frac {1 + \cos 2 \theta}2~d \theta = \frac {25}2 \int (1 + \cos 2 \theta) ~ d \theta

    now what?
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    Quote Originally Posted by Jhevon View Post
    no you didn't. you didn't even integrate correctly

    25 \int \cos^2 \theta ~d \theta = 25 \int \frac {1 + \cos 2 \theta}2~d \theta = \frac {25}2 \int (1 + \cos 2 \theta) ~ d \theta

    now what?
    hmmm I really am not sure. I think what is confusing me is the fact that you have 1 + cos 2 theta. so you have to break the integrals up? and wouldnt this be a u-sub now? with
    u = 2theta du = 2 so put 2 in 1/2 out the integral?

    but then I get 25/4 sin 2theta
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    hmmm I really am not sure. I think what is confusing me is the fact that you have cos 2 theta. wouldnt this be a u-sub now? with
    u = 2theta du = 2 so put 2 in 1/2 out the integral?

    but then I get 25/4 sin 2theta
    yes...u-sub is the hard way. you should know that \int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C, for k \ne 0 a constant.

    thus you get \frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C

    now use your trig sub to back substitute to get your answer in terms of t.
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    Quote Originally Posted by Jhevon View Post
    yes...u-sub is the hard way. you should know that \int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C, for k \ne 0 a constant.

    thus you get \frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C

    now use your trig sub to back substitute to get your answer in terms of t.
    ok another probably obvious question. When you have

    sin 2 theta how would you go about using substitutions when you dont know what theta is? we know that theta = sin^-1(t/5) but how would that even help here?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    ok another probably obvious question. When you have

    sin 2 theta how would you go about using substitutions when you dont know what theta is? we know that theta = sin^-1(t/5) but how would that even help here?
    did you even read the chapter on trig subs? were you paying attention in class?

    since we know \sin \theta = \frac t5 = \frac {\text{opposite}}{\text{hypotenuse}}, we have the following triangle. use it to find \cos \theta = \frac {\text{adjacent}}{\text{hypotenuse}}, and then you can change your answer to be in terms of t. We find the other side by using Pythagoras' theorem
    Attached Thumbnails Attached Thumbnails [SOLVED] trig subs-triangle.png  
    Last edited by Jhevon; April 5th 2009 at 11:55 AM.
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    Quote Originally Posted by Jhevon View Post
    yes...u-sub is the hard way. you should know that \int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C, for k \ne 0 a constant.

    thus you get \frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C

    now use your trig sub to back substitute to get your answer in terms of t.
    Thanks for all your help!!! I have solved my problem.
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  13. #13
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    Quote Originally Posted by Jhevon View Post
    did you even read the chapter on trig subs? were you paying attention in class?

    since we know \sin \theta = \frac t5 = \frac {\text{opposite}}{\text{hypotenuse}}, we have the following triangle. use it to find \cos \theta = \frac {\text{adjacent}}{\text{hypotenuse}}, and then you can change your answer to be in terms of t
    I did read the chapter and have a lot of notes. The trig substitutions section just expects you to know your trig identities which I do not know all of them by heart. I know all the basic identities. The sin2theta I had to look up. Thats what I was confused about.
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