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Thread: Trigonometric Substitution

  1. #1
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    Exclamation Trigonometric Substitution

    Question:
    $\displaystyle \int\frac{2dt}{t\sqrt{t^4+25}}$
    Attempt:
    $\displaystyle \int\frac{2dt}{t\sqrt{(t^2)^2+(5)^2}}$

    a=5 ,$\displaystyle t^2=5tan\theta$,$\displaystyle 2tdt=5sec^2\theta$
    Since there is now differentiation for $\displaystyle sec^2\theta $ can i change it to $\displaystyle tan^2\theta+1$

    Please answer in steps , Thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
    $\displaystyle \int\frac{2dt}{t\sqrt{t^4+25}}$
    Attempt:
    $\displaystyle \int\frac{2dt}{t\sqrt{(t^2)^2+(5)^2}}$

    a=5 ,$\displaystyle t^2=5tan\theta$,$\displaystyle 2tdt=5sec^2\theta$
    Since there is now differentiation for $\displaystyle sec^2\theta $ can i change it to $\displaystyle tan^2\theta+1$

    Please answer in steps , Thank you
    Hello,
    Your substitution works, but from your substitution, clearly $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$, find $\displaystyle dt$ in terms of $\displaystyle \theta$.
    Then change your integral accroding to this substitution; you will get (with some simplification) a simple trigonometric integral.
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  3. #3
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    Quote Originally Posted by General View Post
    Hello,
    Your substitution works, but from your substitution, clearly $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$, find $\displaystyle dt$ in terms of $\displaystyle \theta$.
    Then change your integral accroding to this substitution; you will get (with some simplification) a simple trigonometric integral.
    So ,
    $\displaystyle t^2=5tan\theta$
    $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$
    $\displaystyle dt=\sqrt{5}\,sec(\theta)$
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    So ,
    $\displaystyle t^2=5tan\theta$
    $\displaystyle t=\sqrt{5} \, \sqrt{tan(\theta)}$
    $\displaystyle dt=\sqrt{5}\,\sqrt{sec^2(\theta)}$
    No No No. In general: $\displaystyle \left( \sqrt{f(x)} \right)^{'} \neq \left( \sqrt{f'(x)} \right)$.

    Try to find $\displaystyle dt$ again, and notice that $\displaystyle \sqrt{tan(\theta)} = tan^{\frac{1}{2}}(\theta)$.
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