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Re: MX2 2016 Integration Marathon $\noindent \textbf{Alternatively} -- in finishing the integral off an IBP could be avoided by recognising:$ \begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= 2 \int^{\frac{\pi}{8}}_0 \left (u \sec^2 u + \tan u \right ) \, du\\&= 2...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Find $\int x \cot x \, \mbox{cosec} \, x \, dx.

Re: MX2 2016 Integration Marathon \begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right ] \, dx\\&= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{2 \cos^2 \frac{x}{2}} + \tan \frac{x}{2} \right...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $ \int \frac{x^2 \tan^{-1} x}{1 + x^2} \, dx.

Re: MX2 2016 Integration Marathon $\noindent Note the integral is improper since the integrand is undefined at $x = 0$. Since it converges we can use a clever little boundary switch to avoid having to evaluate the integral directly at this point (in which case one would need to take a limit...

Re: MX2 2016 Integration Marathon Correct and is exactly why I chose the limits of integration I did. The final answer is: 11/6.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Evaluate $\int^1_{-2} \sqrt{x^4 + 2x^3 + x^2} \, dx.

Re: MX2 2016 Integration Marathon $\noindent InteGrand - I think the integrand simplies to $\frac{1 + x^2}{2 + x^2}. $\noindent Note that $\sin (\cot^{-1} x) = \frac{1}{\sqrt{1 + x^2}} = \alpha$ and $\cos (\tan^{-1} \alpha ) = \frac{1}{\sqrt{1 + \alpha^2}}. $\noindent...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Follow-up Question}$ $\noindent Find $\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Evaluate $\int^{\frac{\pi}{2}}_0 \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.

Re: HSC 2016 4U Marathon - Advanced Level $\noindent \textbf{Next Question} $\noindent Let$\\\begin{align*}f(x) &= a_0 x^m + a_1 x^{m - 1} + \cdots + a_m\end{align*}\\$and$\\\begin{align*}g(x) &= b_0 x^n + b_1 x^{n - 1} + \cdots + b_n\end{align*}\\$where $\deg f \leqslant \deg g - 1$ and...

Re: HSC 2016 4U Marathon - Advanced Level Would anyone care to offer up a solution for the benefit of others?

Re: MX2 2016 Integration Marathon Mmmm...seem to be having a particularly bad day at the office. I should have spotted this one a mile off -- 2016ers take note.

Re: MX2 2016 Integration Marathon $\noindent Let $u = \frac{1}{\sqrt{2}} \sin \theta, du = \frac{1}{\sqrt{2}} \cos \theta \, d\theta$. So\\\begin{align*}\int \frac{du}{u^2 \sqrt{1 - 2 u^2}} &= \int \frac{\frac{1}{\sqrt{2}} \cos \theta}{\frac{1}{2} \sin^2 \theta \sqrt{1 - \sin^2 \theta}} \...

Re: HSC 2016 4U Marathon - Advanced Level $\noindent Let $P(x)$ be a polynomial with positive coefficients. Show that if$\\\begin{align*}P \left (\frac{1}{x} \right ) &\geqslant \frac{1}{P(x)}\end{align*}\\$ holds for $x = 1$, then it holds for all $x > 0$.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Find $\int \frac{dx}{\cos (2x) \sin (4x)}.

Re: MX2 2016 Integration Marathon $\noindent First we show $\int \cot^{-1} x \, dx = x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) + \mathcal{C}. $\noindent Using IBP one has\\\begin{align*}\int \cot^{-1} x \, dx &= x \cot^{-1} x - \int x \cdot -\frac{1}{1 + x^2} \, dx = x \cot^{-1} x +...

Re: MX2 2016 Integration Marathon Ah, I just knew there had to be another way.

Re: MX2 2016 Integration Marathon A new question from anyone?

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $ \int^\pi_0 \frac{(x + 3) \sin x}{1 + \cos^2 x} \, dx.