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Math Help - Solving an integral using a special substitution

  1. #1
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    Solving an integral using a special substitution

    Can somebody tell me if this is right?

    I must solve:
    \displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqr  t[ ]{9+x^2}}

    And I used the substitutions:
    x=3sinht
    dx=3coshtdt

    And the identity:

    cosh^2t-sinh^2t=1\Rightarrow{\cosht=\sqrt[ ]{1+sinh^2t}}

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqr  t[ ]{9+x^2}}=\displaystyle\int_{}^{}\displaystyle\frac  {3coshtdt}{\sqrt[ ]{3^2+(3sinht)^2}}=\displaystyle\int_{}^{}\displays  tyle\frac{3coshtdt}{\sqrt[ ]{3^2(1+sinh^2t)}}=\displaystyle\int_{}^{}\displays  tyle\frac{coshtdt}{cosht}=\displaystyle\int_{}^{}d  t=t+C=argsh(\displaystyle\frac{x}{3})+C

    Bye there, and thanks for posting.
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Can somebody tell me if this is right?
    Well done!
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  3. #3
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    Thanks.
    What about this one? I don't know how to proceed now:

    \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin  (t)}

    I've solved this using the table, using the definition of cosecant: \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin  (t)}=\ln|csc(t)-cot(t)|+C

    This was the result of solving: \displaystyle\int_{}^{}\displaystyle\frac{dx}{x\sq  rt[ ]{1-x^2}}=\ln|csc(arcsin(x))-cot(arcsin(x))|+C

    But now I must solve:
    \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\  sqrt[ ]{4-x^2}}

    x=2\sin(t)
    dx=2\cos(t)dt

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\  sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\  cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyl  e\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\dis  playstyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t  )dt

    I thought of solving by parts, but...

    u=csc(t)
    du=-csc(t)cot(t)dt

    dv=csc(t)dt
    v=\ln|csc(t)-cot(t)|
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  4. #4
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    Quote Originally Posted by Ulysses View Post
    What about this one? I don't know how to proceed now:

    \displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin  (t)}
    Do you know the Weierstrass substitution? If If [LaTeX ERROR: Convert failed] , then [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . Using these results we have [LaTeX ERROR: Convert failed]

    But now I must solve:
    \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\  sqrt[ ]{4-x^2}}

    x=2\sin(t)
    dx=2\cos(t)dt

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\  sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\  cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyl  e\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\dis  playstyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t  )dt
    [LaTeX ERROR: Convert failed] is a standard integral. As long you broke down the integral up-to this stage and know this result you should be alright.
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  5. #5
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    Thank you verymuch!
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  6. #6
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    Quote Originally Posted by Ulysses View Post
    Thank you verymuch!
    No problem. I take a delight in helping people who show their work, if I can that is.
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  7. #7
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    Really noble from you sharing your knowledge that way. Believe me, I know that people like you makes this world a better place. I know... a post in a forum doesn't mean much, but is the attitude. And actually the free access to knowledge makes the difference.

    Thanks is all I can give now
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