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Math Help - integral

  1. #1
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    integral

    Find the value of \int \frac{1}{(1+x^2)^{3 \over 2}}dx

    I have been trying to solve this question since many years but I can't reach anywhere...

    I have tried substituting x^2 , x^2+1 and 1/(x^2+1) but I didnot get the answer.
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  2. #2
    Newbie AgentSmith's Avatar
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    Re: integral

    Substitute x= \tan {\lambda}.

    \begin{align*} \int \frac{1}{(1+x^2)^{\frac{3}{2}}} \ dx&= \int \frac{\sec^2{\lambda}}{(\sec^2{\lambda})^{\frac{3}  {2}}}d\lambda \\ &=\int \frac{1}{\sec{\lambda}} \ d\lambda \\ &= \int \cos{\lambda} \ d\lambda \\ &= \sin{\lambda}+C \\ &= \sin[\tan^{-1}{(x)}]+C\end{align*}

    Do you understand now?
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  3. #3
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    Re: integral

    Quote Originally Posted by AgentSmith View Post
    Substitute x= \tan {\lambda}.

    \begin{align*} \int \frac{1}{(1+x^2)^{\frac{3}{2}}} &= \int \frac{\sec^2{\lambda}}{(\sec^2{\lambda})^{\frac{3}  {2}}}d\lambda \\ &=\int \frac{1}{\sec{\lambda}}d\lambda \\ &= \int \cos{\lambda} d\lambda \\ &= \sin{\lambda}+C \\ &= \sin[\tan^{-1}{(x)}]+C\end{align*}

    Do you understand now?
    I most certainly do not understand!
    The answer given in the book is  \frac{x}{\sqrt{x^2+1}}+C ...
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  4. #4
    Newbie AgentSmith's Avatar
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    Re: integral

    Quote Originally Posted by Ozymandias View Post
    I most certainly do not understand!
    The answer given in the book is  \frac{x}{\sqrt{x^2+1}}+C ...
    That is the same thing...

    \sin(\tan^{-1}{x})=\sin{\left [\sin^{-1}{\left(\frac{x}{\sqrt{x^2+1}}}\right) \right]}=\frac{x}{\sqrt{x^2+1}}
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  5. #5
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    Re: integral

    okay! thank you for your explanation
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  6. #6
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    Re: integral

    The reason being...

    \displaystyle \begin{align*} \sin^2{\theta} + \cos^2{\theta} &\equiv 1 \\ \frac{\sin^2{\theta}}{\sin^2{\theta}} + \frac{\cos^2{\theta}}{\sin^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ 1 + \frac{1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{\tan^2{\theta} + 1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{\tan^2{\theta}}{\tan^2{\theta} + 1} &\equiv \sin^2{\theta} \\ \sin{\theta} &\equiv \pm \frac{\tan{\theta}}{\sqrt{\tan^2{\theta} + 1}} \end{align*}

    Therefore

    \displaystyle \begin{align*} \sin{\left(\tan^{-1}{x}\right)} &= \pm \frac{\tan{\left(\tan^{-1}{x}\right)}}{\sqrt{ \tan^2{\left(\tan^{-1}{x}\right) + 1}}} \\ &= \pm \frac{x}{\sqrt{x^2 + 1}} \end{align*}
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