i gave it a shot and this is what i got

= (1/cosx)(1/cosx)(sinx/cosx) (I REWROTE IT)

= (secx)(secxtanx) (TIME TO INTEGRATE)

= (secxtanx)(secx) + C

did i do it right?

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- Dec 13th 2012, 11:10 PMasilvester635Integration problem
i gave it a shot and this is what i got

= (1/cosx)(1/cosx)(sinx/cosx) (I REWROTE IT)

= (secx)(secxtanx) (TIME TO INTEGRATE)

= (secxtanx)(secx) + C

did i do it right? - Dec 13th 2012, 11:43 PMMarkFLRe: Integration problem
No, I think I would approach it with the substitution:

$\displaystyle u=\cos(x)\,\therefore\,du=-\sin(x)\,dx$ and we have:

$\displaystyle -\int u^{-3}\,du=\frac{1}{2u^2}+C=\frac{1}{2\cos^2(x)}+C$

Using your approach, we could let:

$\displaystyle u=\sec(x)\,\therefore\,du=\sec(x)\tan(x)\,dx$ and we have the easier to integrate:

$\displaystyle \int u\,du=\frac{u^2}{2}+C=\frac{\sec^2(x)}{2}+C$

The two results are equivalent.