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Math Help - Integration

  1. #1
    Member srirahulan's Avatar
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    Cool Integration

    \int{\frac{1}{1+cos^2x}d(x)} In this case i divide numerator and denominator by sec^2x and i get \int{\frac{sec^2x}{1+sec^2x}}{ like this.Then What can i do???????
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    Forum Admin topsquark's Avatar
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    Re: Integration

    Quote Originally Posted by srirahulan View Post
    \int{\frac{1}{1+cos^2x}d(x)} In this case i divide numerator and denominator by sec^2x and i get \int{\frac{sec^2x}{1+sec^2x}}{ like this.Then What can i do???????
    Hint: Start over and sub in x = 2u, then use the double angle formula for cos(2u).

    -Dan
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  3. #3
    Member srirahulan's Avatar
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    Re: Integration

    I tried by your way but i have another problem i get 1/3+cos2x d(2x)>>>>
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Integration

    Hmmm.... It would seem that I forgot to square the cos() in the denominator. Sorry!

    -Dan
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  5. #5
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    Re: Integration

    Quote Originally Posted by srirahulan View Post
    \int{\frac{1}{1+cos^2x}d(x)} In this case i divide numerator and denominator by sec^2x and i get \int{\frac{sec^2x}{1+sec^2x}}{ like this.Then What can i do???????
    Well, you'd actually be MULTIPLYING the top and bottom by \displaystyle \begin{align*} \sec^2{(x)} \end{align*} to give you this. Anyway...

    \displaystyle \begin{align*} \int{ \frac{\sec^2{(x)}}{1 + \sec^2{(x)}}\,dx} &= \int{ \frac{\sec^2{(x)}}{ 1 + \left[ 1 + \tan^2{(x)} \right] } \, dx} \\ &= \int{ \frac{\sec^2{(x)}}{ 2 + \tan^2{(x)}}\,dx} \end{align*}

    Now make the substitution \displaystyle \begin{align*} \tan{(x)} = \sqrt{2}\tan{(\theta)} \implies \sec^2{(x)}\,dx &= \sqrt{2}\sec^2{(\theta)}\,d\theta \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{ \frac{\sec^2{(x)}}{2 + \tan^2{(x)}}\,dx} &= \int{ \frac{\sqrt{2}\sec^2{(\theta)}}{2 + \left[ \sqrt{2} \tan{(\theta)} \right] ^2 } \, d\theta} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ 2 + 2\tan^2{(\theta)}}\,d\theta} \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{1 + \tan^2{(\theta)}}\,d\theta} \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta  } \\ &= \frac{\sqrt{2}}{2} \int{ 1\,d\theta} \\ &= \frac{\sqrt{2}}{2} \theta + C \\ &= \frac{\sqrt{2}}{2} \arctan{ \left[ \frac{\sqrt{2}\tan{(\theta)}}{2} \right] } + C  \end{align*}
    Thanks from topsquark and ReneG
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