# First Order Differential Equation Help

• Mar 20th 2011, 09:16 AM
StaryNight
First Order Differential Equation Help
Solving this differential question is supposed to only be worth 6 marks in a past exam question, but my working leads me to a very difficult integral. I was wondering if I have missed something in my solution?

http://latex.codecogs.com/gif.latex?...tan(x))^2%20dx
• Mar 20th 2011, 09:32 AM
TheEmptySet
Quote:

Originally Posted by StaryNight
Solving this differential question is supposed to only be worth 6 marks, but my working leads me to a very difficult integral. I was wondering if I have missed something in my solution?

http://latex.codecogs.com/gif.latex?...tan(x))^2%20dx

No it looks good so far

for the last integral

$\displaystyle \int \sec(x)(\sec(x)+\tan(x))^2dx$

let $u=\sec(x)+\tan(x) \implies du=[\sec(x)\tan(x)+\sec^2(x)]dx \iff du=u\sec(x)dx \iff \frac{du}{u}=\sec(x)dx$

putting this is gives the integral

$\displaystyle \int u^2\frac{du}{u}=\int udu$
• Mar 20th 2011, 09:42 AM
StaryNight
Of course! I should have seen this.
Quote:

Originally Posted by TheEmptySet
No it looks good so far

for the last integral

$\displaystyle \int \sec(x)(\sec(x)+\tan(x))^2dx$

let $u=\sec(x)+\tan(x) \implies du=[\sec(x)\tan(x)+\sec^2(x)]dx \iff du=u\sec(x)dx \iff \frac{du}{u}=\sec(x)dx$

putting this is gives the integral

$\displaystyle \int u^2\frac{du}{u}=\int udu$