# Thread: HSC 2017 MX2 Integration Marathon (archive)

1. ## Re: HSC 2017 MX2 Integration Marathon

$\noindent a) \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 - 1}} \\\\ b) \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 + 1}}$
$a) \quad Let\quad I=\int { \frac { dx }{ x\sqrt { { x }^{ 10 }\pm { x }^{ 5 }+1 } } } \\ Let\quad u={ x }^{ 5 }\\ du=5{ x }^{ 4 }dx\\ I=\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { u^{ 2 }\pm u-1 } } } \\ \quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }-\frac { 5 }{ 4 } } } }\quad$
$Let\quad I=\quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } } } \\ Let\quad w=u\pm \frac { 1 }{ 2 } \\ dw=du\\ I=\frac { 1 }{ 5 } \int { \frac { dw }{ (w\mp \frac { 1 }{ 2 } )\sqrt { { w }^{ 2 }+\frac { 3 }{ 4 } } } } \\ =\frac { 1 }{ 5 } (ln\left| w\mp \frac { 1 }{ 2 } \right| -ln\left| 2\sqrt { { w }^{ 2 }+\frac { 3 }{ 4 } } \pm w+\frac { 3 }{ 2 } \right| )+C\quad \\ =\frac { 1 }{ 5 } (ln\left| u \right| -ln\left| 2\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } \pm u\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ =\frac { 1 }{ 5 } (ln\left| { x }^{ 5 } \right| -ln\left| 2\sqrt { { ({ x }^{ 5 }\pm \frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } } \pm { x }^{ 5 }\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ =\frac { 1 }{ 5 } (ln\left| { x }^{ 5 } \right| -ln\left| 2\sqrt { { x }^{ 10 }\pm { x }^{ 5 }+1 } \pm { x }^{ 5 }\mp \frac { 1 }{ 2 } +\frac { 3 }{ 2 } \right| )+C\\ \\$
$b)\quad Let\quad I=\int { \frac { dx }{ x\sqrt { { x }^{ 10 }\pm { x }^{ 5 }-1 } } } \\ Let\quad u={ x }^{ 5 }\\ du=5{ x }^{ 4 }dx\\ I=\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { u^{ 2 }\pm u-1 } } } \\ \quad =\frac { 1 }{ 5 } \int { \frac { du }{ u\sqrt { { (u\pm \frac { 1 }{ 2 } ) }^{ 2 }-\frac { 5 }{ 4 } } } } \\ Let\quad w=u\pm \frac { 1 }{ 2 } \\ dw=du\\ I=\frac { 1 }{ 5 } \int { \frac { dw }{ (w\mp \frac { 1 }{ 2 } )\sqrt { { w }^{ 2 }-\frac { 5 }{ 4 } } } } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { \frac { w }{ 2 } \mp \frac { 5 }{ 4 } }{ \sqrt { { w }^{ 2 }-\frac { 5 }{ 4 } } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { \frac { u }{ 2 } \pm \frac { 1 }{ 4 } \mp \frac { 5 }{ 4 } }{ \sqrt { { u }^{ 2 }\pm u-1 } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { u\pm 2 }{ 2\sqrt { { u }^{ 2 }\pm u-1 } } )+C } \\ I=\pm \frac { 1 }{ 5 } \tan ^{ -1 }{ (\frac { { x }^{ 5 }\mp 2 }{ 2\sqrt { { x }^{ 10 }\pm { x }^{ 5 }-1 } } )+C }$
though in reality these questions are trivial and can easily be done by inspection

2. ## Re: HSC 2017 MX2 Integration Marathon

If b>a>0 evaluate:

$\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x$

3. ## Re: HSC 2017 MX2 Integration Marathon

If b>a>0 evaluate:

$\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x$
Trivial by inspection

$\noindent I=\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\tan^{-1}{ax} - \cot^{-1}{bx}}{x} \, \text{d}x \\\\ Let x=\dfrac{1}{y} \\\\ \text{d}x = \dfrac{-\text{d}y}{y^2} \\\\ I=\int _b^a \dfrac{\tan^{-1}{\tfrac{a}{y}}-\cot^{-1}{\tfrac{b}{y}}}{\tfrac{1}{y}} \dfrac{-\text{d}y}{y^2} \\\\ I =\int_a^b \dfrac {\tan^{-1}{\tfrac{a}{y}}- \cot^{-1}{\tfrac{b}{y}}}{y} \text{d}y \\\\ I =\int_a^b \dfrac{\cot^{-1}{\tfrac{y}{a}} - \tan^{-1}{\tfrac{y}{b}}}{y} \text{d}y \\\\ Let y=abx \\\\I =\int_{\tfrac{1}{b}}^{\tfrac{1}{a}} \dfrac{\cot^{-1}{bx} - \tan^{-1}{ax}}{abx} \text{d}x \\\\ \quad =-ab\int _{ \tfrac{1}{b}}^{ \tfrac{1}{a}} \frac{\tan^{-1}{ax}-\cot^{-1}{bx}}{x} \text{d}x \\\\ \quad I=-abI\\\\ \therefore \quad I=0 since ab > 0 > -1$
Note the potential tautology if ab=-1 cannot occur as b>a>0.

4. ## Re: HSC 4U Integration Marathon 2017

$\noindent \int \frac{\sqrt{\cos{2x}}}{\sin{x}} \text{d}x = \int \frac{\sqrt{2\cos^2{x}-1}}{1-\cos^2{x}} \sin{x} \text{d}x \\\\ \sqrt{2} \cos{x} = \sec{\theta} \Rightarrow -\sqrt{2} \sin{x} \text{d}x = \sec{\theta} \tan{\theta} \text{d}\theta \\\\ \sqrt{2}\int \frac{\sqrt{(\sqrt{2} \cos{x})^2 -1}}{(2\cos{x})^2 -2} (-\sqrt{2} \sin{x} \text{d}x) \\\\ = \sqrt{2} \int \frac{\sec{\theta} \tan^2{\theta}}{\sec^2{\theta} -2} \text{d}\theta = \sqrt{2} \int \frac{\sin^2{\theta} \cos{\theta} \, \, \text{d}(\theta)}{(1-\sin^2{\theta}) (2\sin^2{\theta}-1)} \\\\ \sin^2{\theta} \equiv (2\sin^2{\theta}-1) + (1-\sin^2{\theta}) \\\\ \Rightarrow \frac{\sin^2{\theta}}{(1-\sin^2{\theta})(2\sin^2{\theta}-1)} \equiv \sec^2{\theta} + \frac{1}{2\sin^2{\theta}-1} \\\\ \sqrt{2}\int \sec{\theta} \, \text{d}\theta - \sqrt{2} \int \frac{\text{d}(\sin{\theta})}{1-2\sin^2{\theta}}$

$\noindent = \sqrt{2} \log{(\sec{\theta} + \tan{\theta})} + \frac{1}{2} \log{\left(\frac{1-\sqrt{2}\sin{\theta}}{1+\sqrt{2}\sin{\theta}} \right)} + \mathcal{C} \\\\ \sec{\theta} = \sqrt{2}\cos{x},\, \tan{\theta} = \sqrt{\cos{2x}}, \sin{\theta} = \frac{\sqrt{\cos{2x}}}{\sqrt{2} \cos{x}} \\\\ \sqrt{2} \log(\sqrt{2}\cos{x} + \sqrt{\cos{2x}}) + \frac{1}{2} \log{\left( \frac{\cos{x} - \sqrt{\cos{2x}}}{\cos{x} + \sqrt{\cos{2x}}} \right)} + \mathcal{C}$

Simplifying the derivative in Mathematica confirms the result.
Hey, sorry for being so bad at this subject, but how do you know to use that substitution? Thinking process? What are some key things to look out for when attempting a question?

5. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by EngineeringHelp
Hey, sorry for being so bad at this subject, but how do you know to use that substitution? Thinking process? What are some key things to look out for when attempting a question?
Eliminate irrational functions using whatever you have at your disposal.

The only three ways of doing so are:

Classical Trigonometric Substitution (High School level)

Hyperbolic Trigonometric Substitution (University level)

Euler Substitution (Extracurricular)

6. ## Re: HSC 4U Integration Marathon 2017

Eliminate irrational functions using whatever you have at your disposal.

The only three ways of doing so are:

Classical Trigonometric Substitution (High School level)

Hyperbolic Trigonometric Substitution (University level)

Euler Substitution (Extracurricular)
Ooh, so for the HSC all i should worry about is a Trig Sub? As it wouldn't be right to be asked questions requiring the other two, right?

7. ## Re: HSC 4U Integration Marathon 2017

Also, does anyone have any 'not so complicated' reduction-formula questions? If so, please post. Im assuming Paradoxica will be supplying solutions xD <3

8. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by EngineeringHelp
Also, does anyone have any 'not so complicated' reduction-formula questions? If so, please post. Im assuming Paradoxica will be supplying solutions xD <3
http://prntscr.com/fnxmsv

9. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
$\int_0^1 \frac{x^n}{\sqrt{x+1}} \, \text{d}x$
...ew

10. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by pikachu975
Lmao use Mathematical Induction to prove? Stuff In

11. ## Re: HSC 2017 MX2 Integration Marathon

\begin{align*}I_n&=\int_0^1\left(\frac{x^n+x^{n-1}-x^{n-1}}{\sqrt{x+1}}\right)dx\\ &= \int_0^1 \frac{x^{n-1}(x+1)}{\sqrt{x+1}}-I_{n-1}\\ &\stackrel{IBP}{=}\frac{x^n}n \sqrt{x+1} \bigg|_0^1 - \frac{1}{2n}\int_0^1\frac{x^n}{\sqrt{x+1}}-I_{n-1}\\ \implies \left(1+\frac{1}{2n}\right)I_n&=\frac{\sqrt2}{n} - I_{n-1}\\ (2n+1)I_n&=2\sqrt2 - 2nI_{n-1} \end{align*}

12. ## Re: HSC 2017 MX2 Integration Marathon

Try this one.
$\cot 3x\sec^2(ln \sqrt{sin 3x})$

13. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by stupid_girl
Try this one.
$\cot 3x\sec^2(ln \sqrt{sin 3x})$
d(lnsqrt(sin3x)) = 3/2 * cot3x * dx
2/3 * d(lnsqrt(sin3x)) = cot3x dx

Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C

14. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by pikachu975
d(lnsqrt(sin3x)) = 3/2 * cot3x * dx
2/3 * d(lnsqrt(sin3x)) = cot3x dx

Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C

15. ## Re: HSC 2017 MX2 Integration Marathon

long time no integrals
here's a couple of good ones to try out!

$\noindent \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )} dx \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2 dx \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}} dx \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right ) dx$

16. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by jathu123
long time no integrals
here's a couple of good ones to try out!

$\noindent \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )} dx \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2 dx \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}} dx \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right ) dx$
$\noindent 1) By inspection \frac{\left(x^2-2\right)}{\left(x^4+5x^2+4\right)} is the derivative of tan^{-1}\left(x+\frac{2}{x}\right) hence the integral becomes: \\ \\ \quad \frac{d\left(tan^{-1}\left(x+\frac{2}{x}\right)\right)}{tan^{-1}\left(x+\frac{2}{x}\right)} which equals ln\left|tan^{-1}\left(x+\frac{2}{x}\right)\right|+C$

$\noindent 4) let \quad u = lnx , \quad e^u du = dx \\ Integral becomes \int \:\left(\frac{1}{u}+lnu\right)e^udu and applying reverse product rule you get \quad xln(lnx) + C.$

17. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by jathu123
long time no integrals
here's a couple of good ones to try out!

$\noindent \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )} dx \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2 dx \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}} dx \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right ) dx$
$\noindent The second one can also be found using the reverse product rule. Begin by noting that$

\begin{align*}\left (\frac{1 - x}{1 + x^2} \right )^2 &= \frac{(1 - x)^2}{(1 + x^2)^2} = \frac{1 + x^2}{(1 + x^2)^2} - \frac{2x}{(1 + x^2)^2} = \frac{1}{1 + x^2} - \frac{2x}{(1 + x^2)^2} = f(x) + f'(x),\end{align*}

$\noindent where f(x) = \frac{1}{1 + x^2}.$

$\noindent So the integral becomes$

\begin{align*}\int e^x \left (\frac{1 - x}{1 + x^2} \right )^2 \, dx &= \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C = \frac{e^x}{1 + x^2} + C.\end{align*}

18. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by jathu123
long time no integrals
here's a couple of good ones to try out!

$\noindent \int \frac{x^2-2}{\left ( x^4+5x^2+4 \right )\tan^{-1}\left ( x+\frac{2}{x} \right )} dx \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2 dx \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}} dx \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right ) dx$
$\noindent There is quite a nice way to do the third one. We begin by noting that$

\begin{align*}\sin 2x &= 2 \sin x \cos x = 1 + 2 \sin x \cos x - 1 = (\cos^2 x + \sin^2 x) + 2\sin x \cos x - 1 = (\cos x + \sin x)^2 - 1.\end{align*}

$\noindent So, for the integral we have$

\begin{align*}\int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx &= \int \frac{\sin x - \cos x}{\sqrt{(\cos x + \sin x)^2 - 1}} \, dx.\end{align*}

$\noindent Letting u = \cos x + \sin x, du = -(\sin x - \cos x) \, dx so$

\begin{align*}\int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx &= -\int \frac{du}{\sqrt{u^2 - 1}}\\ &=-\ln |u + \sqrt{u^2 - 1}| + C\\ &= -\ln |\cos x + \sin x + \sqrt{\sin 2x}| + C.\end{align*}

19. ## Re: HSC 2017 MX2 Integration Marathon

Integrate 1/[(sinx)sqrt(sin2x)]

20. ## Re: HSC 2017 MX2 Integration Marathon

The new 2018 marathon has been created.

dan

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