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Re: HSC 2018 MX2 Marathon This one is not that too tough, right? T_1=2\sqrt{T_1}-1 $Let $x=\sqrt{T_1} x^2=2x-1 (x-1)^2=0 x=1 T_1=x^2=1^2 $The statement is true for $n=1$. $Assume the statement is true for $n=k$. T_{k+1}=2\sqrt{T_1+T_2+...+T_k+T_{k+1}}-1 $Let...

A few new integrals If you can solve one of them, then you can probably solve all of them. \int\left(x+\sqrt{x^2+1}\right)^{\sqrt{2019}}dx \int\left(\sqrt{x^2+1}-x\right)^{\pi}}dx \int\left(\sec x+\tan x\right)^{2019}\sec^2xdx \int\left(\cot x+\csc x\right)^{\frac{22}{7}}\csc^2xdx

Using the above substitution, it should be obvious that \int\left(\tan^{-1}\sqrt{2x-1}\right)\left(2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}\right)^kdx =\frac{\left(2x\tan^{-1}\left(\sqrt{2x-1}\right)-\sqrt{2x-1}\right)^{k+1}}{2\left(k+1\right)}+c$ for $k\neq-1 The definite integral can be evaluated...

Hint: You may consider the following substitution. u=2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}

This is a new one. \int_1^2\frac{\tan^{-1}\sqrt{2x-1}}{\left(2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}\right)^3}dx Once again, the answer looks quite ugly. \frac{1}{\left(\pi-2\right)^2}-\frac{9}{4\left(4\pi-3\sqrt{3}\right)^2}

As usual, reverse quotient rule problems are easy to set but difficult to solve. It becomes a piece of cake IF (a big if) you can spot it.:) \int\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx=-\frac{\left(1+\cos\ x\right)^2}{4\left(x-\sin x\right)^2}+c

A new one \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx The answer looks quite ugly and probably can't be simplified further. Taking common denominator and expanding out will make it really messy...

This is a new one. Feel free to share your attempt. \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\tan^4x+\cot^4x}{1+\tan^{\frac{2019}{2020}}x}dx

The answer for this one is \frac{3}{\ln2}

Re: HSC 2018 MX2 Integration Marathon The answer for this one is 25+\frac{3}{2}\ln2-\frac{17}{\ln2}

Re: HSC 2018 MX2 Integration Marathon The answer for this one is \frac{25\sqrt{3}}{36}

Re: HSC 2018 MX2 Marathon $\noindent Let $T_n$ be a sequence such that $T_1=1$, $T_2=2$ and $T_{n+2}=T_{n+1}+T_n$. Show by mathematical induction that $T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n$.

Re: HSC 2018 MX2 Marathon $\noindent Let $T_n$ be a positive sequence such that $T_n=2\sqrt{T_1+T_2+...+T_n}-1$. Show by mathematical induction that $T_1+T_2+...+T_n=n^2$.

#83 and #88 are still outstanding and this is a new one. Feel free to share your attempt. \int_0^1\left(\sqrt{4-4^x}\sqrt{7\left(16^x\right)+16^{2x}-5\left(4^{3x}\right)}+\sqrt[3]{2^{10x}-2^{8x+1}}\right)dx

Re: HSC 2018 MX2 Integration Marathon This is slightly tedious. \int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} -...

Re: HSC 2018 MX2 Integration Marathon I saw another approach on the internet...however the back substitution may be slightly messier.

Re: HSC 2018 MX2 Integration Marathon Not sure if anyone attempted to go further from this. If you are careful with the manipulation, you should have got the final answer. \int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx...

Re: HSC 2018 MX2 Integration Marathon This is quite similar to the other beast. \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx =\int_0^{\frac{\pi}{3}}(\sec x)\sqrt{3+\cos2x}dx =\sqrt{2}\int_0^{\frac{\pi}{3}}\frac{\cos x}{\sqrt{2-\sin^2...

Re: HSC 2018 MX2 Integration Marathon This is one possible answer. 2\sqrt{2}\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor + (-1)^{\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor}(\sin x-\cos x)+c By exploiting the periodicity of tan-1(tan x), it can also be expressed as...

Re: HSC 2018 MX2 Integration Marathon This one requires the same trick you've seen. \int_{-\frac{3}{4}}^{\frac{4}{3}}\frac{1}{(1+2\cos^2(\pi x))}dx