The question goes:

Expand $(1-2x)^{1/2}-(1-3x)^{2/3}$ as far as the 4th term.

Ans: $x + x^2/2 + 5x^3/6 + 41x^4/24$

How should I do it?
['https://math.stackexchange.com/questions/49182', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/11103/']
You can translate the inequality to $\frac{y^2 \cdot y'}{1+y^6} \leq x^{2011}$ and then integrate from $x\_00$ to $s$ with respect to $x$. 

$$ \int\_{x\_0}^s \frac{y(x)^2 \cdot y(x)'}{1+y(x)^6} dx \leq \frac{s^{2012}}{2012}-\frac{x\_0^{2012}}{2012}$$

$$ \frac{1}{3} \arctan y^3(s) -\frac{1}{3}\arctan y^3(x\_0) \leq \frac{s^{2012}}{2012}-\frac{x\_0^{2012}}{2012}$$ 

In your case, take $x\_0=\frac{\pi}{4}$
What you basically have is this:

\begin{align\*}
\frac{dy}{dx} & \leq \frac{x^{2011}\cdot (1+y^{6})}{y^{2}} \\ \Longrightarrow \int\frac{y^{2}}{1+y^{6}}\ dy &\leq \int x^{2011} \ dx
\end{align\*}