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Math Help - Integration by Trig Substition

  1. #1
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    Integration by Trig Substition

    Lower boundary = 2sqrt2

    upper boundary = 4

    (sqrt(x^2 - 4)) / x

    So far i've set x = 2 sectheta

    dx = 2 (sec theta) (tan theta)


    From here i'm having problems what I should do next

    I know I have to place new boundaries.

    The new lower boundary is gonna pi / 4

    the upper boundary i'm having trouble figuring out. what is the sec theta when it equals 2?
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  2. #2
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    Quote Originally Posted by JonathanEyoon View Post
    Lower boundary = 2sqrt2

    upper boundary = 4

    (sqrt(x^2 - 4)) / x

    So far i've set x = 2 sectheta

    dx = 2 (sec theta) (tan theta)


    From here i'm having problems what I should do next

    I know I have to place new boundaries.

    The new lower boundary is gonna pi / 4

    the upper boundary i'm having trouble figuring out. what is the sec theta when it equals 2?
    4=2\sec(\theta) \iff 4=\frac{2}{\cos(\theta)} \iff \cos(\theta)=\frac{1}{2} \iff \theta =\cos^{-1}\left( \frac{1}{2}\right)=\frac{\pi}{3}
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  3. #3
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    You have the correct substitution.

    Subbing in you get:

    \int\frac{\sqrt{(2sec(t))^{2}-4}}{2sec(t)}2sec(t)tan(t)dt

    This all reduces down to:

    2\int{tan^{2}(t)}dt

    Now, the new limits are 2\sqrt{2}=2sec(t)\rightarrow{t=\frac{\pi}{4}}

    4=2sec(t)\rightarrow{t=\frac{\pi}{3}}

    So, we have:

    2\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}tan^{2}(t)dt

    Can you finish now?.
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  4. #4
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    I appreciate it fellas. I'll take it from here and let you know if I have any other troubles. Thanks
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  5. #5
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    Actually I am gonna need some further help. How would I integrate tan^2(x)?


    I'll go ahead and take a guess and say. Break up the tan^2(X)

    sin^2(x) * (1 / cos^2(x))

    U - substitution?
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  6. #6
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    Quote Originally Posted by JonathanEyoon View Post
    Actually I am gonna need some further help. How would I integrate tan^2(x)?


    I'll go ahead and take a guess and say. Break up the tan^2(X)

    sin^2(x) * (1 / cos^2(x))

    Integration by parts?
    Try this trig identity

    \tan^2(x)+1=\sec^2(x) \iff \tan^{2}(x)=\sec^2(x)-1

    remember that \frac{d}{dx}\tan(x)=\sec^2(x)
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  7. #7
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    I tried to use that identity but unfortunately i'm just getting more confused. Can you give me a hint on what the next step would be? Thanks


    Never mind. I figured it out.!! *sigh* I hate it when I make careless mistakes!
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  8. #8
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    Quote Originally Posted by JonathanEyoon View Post
    I tried to use that identity but unfortunately i'm just getting more confused. Can you give me a hint on what the next step would be? Thanks

    \int \tan^2(x)dx=\int(\sec^2(x)-1)dx

    from my hint before that \frac{d}{dx}\tan(x)=\sec^2(x) this implies that \int \sec^2(x)dx=\tan(x) so we get...

    \int \tan^2(x)dx=\int(\sec^2(x)-1)dx=\tan(x)-x
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  9. #9
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    Hello,

    Without using sec..

    \tan^2x=\underbrace{1+\tan^2x}_{\text{derivative of tan x}}-1

    --> \int \tan^2x \ dx=\int 1+\tan^2x \ dx-\int dx

    Just another way to do it
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