Hi all.

I started with this equation:

$\displaystyle \frac {dy}{dx}=tan^2(x+y)$

I made the substitution $\displaystyle u = x+y$ and am now looking at this:

$\displaystyle \frac {du}{tan^2u+1} = dx$

How the heck do I integrate that left side?

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- Jun 18th 2012, 04:55 AMphys251Made the substitution, but now how to integrate?
Hi all.

I started with this equation:

$\displaystyle \frac {dy}{dx}=tan^2(x+y)$

I made the substitution $\displaystyle u = x+y$ and am now looking at this:

$\displaystyle \frac {du}{tan^2u+1} = dx$

How the heck do I integrate that left side? - Jun 18th 2012, 07:35 AMBobPRe: Made the substitution, but now how to integrate?
$\displaystyle 1+\tan^{2}A=\sec^{2}A,$

$\displaystyle \frac{1}{\sec^{2}A}=\cos^{2}A=\frac{1}{2}(1+\cos2A ).$ - Jun 19th 2012, 05:22 AMphys251Re: Made the substitution, but now how to integrate?
Oh! Durr, trig identities! Thanks.