anti derivative please

**sluggerbroth** · July 13th 2012, 05:28 AM

sin y dy/ cos^2 y

**Prove It** · July 13th 2012, 05:43 AM

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

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

$\displaystyle \begin{align*} -\int{\frac{1}{u^2}\,du} &= -\int{u^{-2}\,du} \\ &= -\left(\frac{u^{-1}}{-1}\right) + C \\ &= \frac{1}{u} + C \\ &= \frac{1}{\cos{y}} + C \\ &= \sec{y} + C \end{align*}$

**skeeter** · July 13th 2012, 06:31 AM

Originally Posted by sluggerbroth

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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