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09-05-2007

I'll try one of the easy ones for warmup; I'll do the rest, if I can, later.

8) This is quite easy. You just break it into sections, and use the product rule.

$\int sec^2(\theta)tan^2(\theta) d(\theta)$

So, first work with
$\int\sec^2{\theta}$
Is equal to
$\int \ln{|sec{\theta}+\tan{\theta}|}^2 \text{ d}(\theta)$
as every good calc student knows. (Well, plus a constant, but we'll just stick that in at the end)
now, $\int\tan^2{\theta}\text{ d}(\theta)$ is equal to
$-\ln{|cos x|}^2$

Now apply the product rule
$\int f(x)\cdot g(x)=(\int f(x))g(x)+f(x)(\int g(x))$

I'll leave you to solve.

09-05-2007	#5
Temperal Guest Posts: n/a	I'll try one of the easy ones for warmup; I'll do the rest, if I can, later. 8) This is quite easy. You just break it into sections, and use the product rule. $\int sec^2(\theta)tan^2(\theta) d(\theta)$ So, first work with $\int\sec^2{\theta}$ Is equal to $\int \ln{\|sec{\theta}+\tan{\theta}\|}^2 \text{ d}(\theta)$ as every good calc student knows. (Well, plus a constant, but we'll just stick that in at the end) now, $\int\tan^2{\theta}\text{ d}(\theta)$ is equal to $-\ln{\|cos x\|}^2$ Now apply the product rule $\int f(x)\cdot g(x)=(\int f(x))g(x)+f(x)(\int g(x))$ I'll leave you to solve. Last edited by Temperal; 09-05-2007 at 08:33 PM..