Attachment 26214

sorry it's not equal to zero, I was looking at the wrong answer:

I tried using the f'(t)h(f(t) - g'(t)h(g(t)) method but I got

$\displaystyle 2sinx cosx(csc^2x+sec^x)$

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- Dec 13th 2012, 05:51 PMkingsolomonsgravewhy is this equal to zero?
Attachment 26214

sorry it's not equal to zero, I was looking at the wrong answer:

I tried using the f'(t)h(f(t) - g'(t)h(g(t)) method but I got

$\displaystyle 2sinx cosx(csc^2x+sec^x)$ - Dec 15th 2012, 11:49 PMx3bnmRe: why is this equal to zero?
You can solve it like this:

$\displaystyle \begin{align*} \frac{d}{dy}\int_{\cos^{2}(y)}^{\sin^{2}(y)}\frac{ 1}{t} dt =& \frac{d}{dy}\left( \ln(t) \Big{|}^{\sin^{2}(y)}_{\cos^{2}(y)}\right) \\ =& \frac{d}{dy}[\ln{(\sin^{2}(y))} - \ln{(\cos^{2}(y))}] \\ =& \frac{2 \sin(y) \cos(y)}{\sin^{2}(y)} + \frac{2 \sin(y) \cos(y)}{\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos^{3}(y) + 2\sin^{3}(y)\cos(y)}{\sin^{2}(y)\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos(y)(\sin^{2}(y) + \cos^{2}(y))}{\sin^{2}(y)\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos(y)}{\sin^{2}(y)\cos^{2}(y)} \\ =& 2\csc(y)\sec(y) \end{align*}$