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

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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

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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

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

Of course! I should have seen this.

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Originally Posted by TheEmptySet

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

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

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