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Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Evaluate $\int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, dx

Re: MX2 2016 Integration Marathon $\noindent I'll do one of these. The rest are done in a similar fashion which I'll save for others to try. Consider the third one (c). Noting that$ n \cos (n + 1) x \sin^{n - 1} x = n \cos nx \cos x \sin^{n - 1} x - n \sin nx \sin^n x = \frac{d}{dx} (\cos...

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int\frac{2x + 5}{x(x + 1)(x + 4)(x+5) + 5} \, dx.

Re: MX2 2016 Integration Marathon $\noindent We shall solve this one using what I call the {\em reverse quotient rule}. $\noindent As is well known, from the quotient rule we have $\left (\frac{f}{g} \right )' = \frac{f' g - g' f}{g^2}, \,\, g \neq 0. $\noindent We now try to write the...

Re: MX2 2016 Integration Marathon $\noindent Here is a more direct method which avoids the $\int^1_0 \frac{\ln (x + 1)}{x^2 + 1} \, dx$ integral. $\noindent Let $u = \frac{1 - x}{1 + x} \Rightarrow x = \frac{1 - u}{u + 1}, dx = -\frac{2}{(u + 1)^2} \, du$. For the limits, when $x = 1, u =...

Why on Earth would you want to do such a thing? Micro$oft Word will not "mark it up" or "LaTeX" it for you. Instead it will just show you the source code you have typed.

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question} $\noindent Evaluate $\int^1_0 \frac{\tan^{-1} x}{x + 1}.

Re: MX2 2016 Integration Marathon $\noindent Consider the integral $I= \int^1_{-a} \frac{x^2 e^{\tan^{-1} x}}{\sqrt{x^2 + 1}} \, dx,$ where $0 \leqslant a \leqslant 1. $\noindent Let $x = \tan u, dx = \sec^2 u \, du.$ Limits: $ x = 1, u = \pi/4$ and $x = -a, u = -\tan^{-1}(a) = \alpha.$...

Text must also be bounded by dollar signs (that is all text must be enclosed with a dollar sign at the beginning and the end). Also, I would recommend that if you want to see how someone produced the text they have written, just click on the reply button and you can clearly see what they have...

Re: MX2 2016 Integration Marathon $\noindent For those who lack Paradoxica's insight into this one (or though I must admit his comment \textit{by inspection} did make me laugh) I will aim to find the integral using a symmetric substitution of the form $u = x + \frac{1}{x}. $\noindent...

Re: MX2 2016 Integration Marathon $\noindent One does. What one needs to do is try and express the integrand in a form which can be suitably integrated in elementary terms. It's not easy, and would require significant hand holding if it ever appeared in an HSC exam (which I doubt), but it is...

Re: MX2 2016 Integration Marathon $\noindent$\ldots$ and as a \textit{bonus}, find $\int \frac{\cos nx - \cos ny}{\cos x - \cos y} \, dx$ for $n = 2,3,\ldots

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent (a) [easy] Evaluate $\int^\pi_0 \frac{\cos (2x) - \cos (2y)}{\cos x - \cos y} \, dx $\noindent (b) [hard] Evaluate $\int^\pi_0 \frac{\cos (nx) - \cos (ny)}{\cos x - \cos y} \, dx,$ where $n = 0,1,2,\ldots

Re: MX2 2016 Integration Marathon $\noindent Let $I = \int_{-a}^{a} \frac{\text{d}x}{1+x^5+\sqrt{1+x^{10}}}. $\noindent Using the result $\int^b_a f(x) \, dx = \int^b_a f(b + a - x) \, dx$, which for symmetric limits is the same thing as making a substitution of $x = -u$, we have$ I =...

Unfortunately for the Sciences it's sad but true.

Because as a name we all know 4 Unit is so such better than Extension 2 :)

Re: MX2 2016 Integration Marathon $\noindent \textbf{Next Question}$ $\noindent Find $\int \frac{dx}{(x + 2)^{\frac{6}{5}} (x - 7)^{\frac{4}{5}}}.

Re: MX2 2016 Integration Marathon $\noindent Need to keep the train moving right along with a \textbf{New Question}$ $\noindent Evaluate $\int^1_0 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, dx.

Re: MX2 2016 Integration Marathon $\noindent Ah yes, the second one just equals $\sin^{-1} x$. So for completeness$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2 + 1} \, dx = \sqrt{2} \tan^{-1} \left ( \frac{\sqrt{2} x}{\sqrt{1 - x^2}} \right ) + \sin^{-1} x + \cal{C}\end{align*}

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