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Math Help - anti derivative please

  1. #1
    Member sluggerbroth's Avatar
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    anti derivative please

    sin y dy/ cos^2 y
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  2. #2
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    Re: anti derivative please

    \displaystyle \begin{align*} \int{\frac{\sin{y}}{\cos^2{y}}\,dy} = -\int{\frac{-\sin{y}}{\cos^2{y}}\,dy} \end{align*}

    Let \displaystyle \begin{align*} u = \cos{y} \implies du = -\sin{y}\,dy \end{align*} and the integral becomes

    \displaystyle \begin{align*} -\int{\frac{1}{u^2}\,du} &= -\int{u^{-2}\,du} \\ &= -\left(\frac{u^{-1}}{-1}\right) + C \\ &= \frac{1}{u} + C \\ &= \frac{1}{\cos{y}} + C \\ &= \sec{y} + C \end{align*}
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  3. #3
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    Re: anti derivative please

    Quote Originally Posted by sluggerbroth View Post
    sin y dy/ cos^2 y
    ever wonder why you had to learn all those identities ?

    \frac{\sin{y}}{\cos^2{y}} = \frac{1}{\cos{y}} \cdot \frac{\sin{y}}{\cos{y}} = \sec{y} \cdot \tan{y}

    \int \sec{y} \cdot \tan{y} \, dy = \sec{y} + C
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