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?
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?
No it looks good so far
for the last integral
$\displaystyle \displaystyle \int \sec(x)(\sec(x)+\tan(x))^2dx$
let $\displaystyle u=\sec(x)+\tan(x) \implies du=[\sec(x)\tan(x)+\sec^2(x)]dx \iff du=u\sec(x)dx \iff \frac{du}{u}=\sec(x)dx$
putting this is gives the integral
$\displaystyle \displaystyle \int u^2\frac{du}{u}=\int udu$