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Math Help - Trigonometric Substitution

  1. #1
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    Exclamation Trigonometric Substitution

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

    a=5 , t^2=5tan\theta, 2tdt=5sec^2\theta
    Since there is now differentiation for  sec^2\theta can i change it to  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:
    \int\frac{2dt}{t\sqrt{t^4+25}}
    Attempt:
    \int\frac{2dt}{t\sqrt{(t^2)^2+(5)^2}}

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

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

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