HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

omegadot · Nov 27, 2015

Re: MX2 2016 Integration Marathon

kawaiipotato said:
$New Question$

$$ Show $ \int_{0}^{a} \frac{x}{\cos x \cos (a-x)} dx = \frac{a}{\sin a }\ln \sec a$

$$\noindent The integral does not converge for all values of $ a.$ For convergence I will assume $-\frac{\pi}{2} < a < \frac{\pi}{2}.$

$\noindent Let $I = \int^a_0 \frac{x}{\cos x \cos (a - x)} \, dx.$ Using the well-known result of$

$\int^a_0 f(x) \, dx = \int^a_0 f(a - x) \, dx,$ one has$$

$\begin{align*}I &= \int^a_0 \frac{a - x}{\cos (a - x) \cos x} \, dx\\&= a \int^a_0 \frac{dx}{\cos x \cos (a - x)} - I\\\Rightarrow 2 I &= a \int^a_0 \frac{dx}{\cos x \cos (a - x)}\\\Rightarrow I &= a \int^a_0 \frac{dx}{2 \cos x \cos (a - x)}\end{align*}$

$$\noindent Applying the appropriate prosthaphaeresis formula to the denominator gives$\\\begin{align*}I &= a \int^a_0 \frac{dx}{\cos a + \cos (2x - a)}\end{align*}$

$$\noindent Let $2u = 2x - a, du = dx, u \in \left (-\frac{a}{2},\frac{a}{2} \right )$. One therefore has$

$\begin{align*}I &= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{du}{\cos a + \cos 2u}\\&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{du}{\cos a - 1 + 2 \cos^2 u}\\&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sec^2 u}{(\cos a - 1) \sec^2 u + 2} \, du&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sec^2 u}{(\cos a + 1) + (\cos a - 1)\tan^2 u} \, du.\end{align*}$

$$\noindent Now let $t = \tan u, dt = \sec^2 u \, du, t \in \left (-\tan \frac{a}{2},\tan \frac{a}{2} \right ) = (-\alpha,\alpha)$ where $\alpha = \tan \frac{a}{2}.$ Thus$

$\begin{align*}I &= a \int^\alpha_{-\alpha} \frac{dt}{(\cos a + 1) + (\cos a - 1) t^2}\end{align*}$

$$\noindent We now note that as $0 < \cos a \leqslant 1$ for all $-\frac{\pi}{2} < a < \frac{\pi}{2}$ then $\cos a + 1 > 0$ and $\cos a - 1 \leqslant 0$. So$

$\begin{align*}I &= a \int^\alpha_{-\alpha} \frac{dt}{(\cos a + 1) - (1 - \cos a) t^2}\\&= \frac{a}{1 - \cos a} \int^\alpha_{-\alpha} \frac{dt}{\frac{1 + \cos a}{1 - \cos a} - t^2}\\&= \frac{a}{1 - \cos a} \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{dt}{\cot^2 \frac{a}{2} - t^2}, \quad \mbox{since} \,\, \cot^2 \frac{a}{2} = \frac{1 + \cos a}{1 - \cos a}\\&= \frac{a}{1 - \cos a} \cdot \frac{1}{2 \cot \frac{a}{2}} \left [\ln \left |\frac{\cot \frac{a}{2} + t}{\cot \frac{a}{2} - t} \right | \right ]^\alpha_{-\alpha}\end{align*}$

$$\noindent Here the result $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left |\frac{a + x}{a - x} \right | + \mathcal{C}$ has been used$

$\begin{align*}I &= \frac{a}{1 - \cos a} \frac{1}{2} \sqrt{\frac{1 - \cos a}{1 + \cos a}} \left [\ln \left |\frac{\cot \frac{a}{2} + \tan \frac{a}{2}}{\cot \frac{x}{2} - \tan \frac{a}{2}} \right | - \ln \left |\frac{\cot \frac{a}{2} - \tan \frac{a}{2}}{\cot \frac{x}{2} + \tan \frac{a}{2}} \right | \right ]\\&= \frac{a}{\sqrt{1 - \cos^2 a}} \ln \left |\frac{\cot \frac{a}{2} + \tan \frac{a}{2}}{\cot \frac{x}{2} - \tan \frac{a}{2}} \right |\\&= \frac{a}{|\sin a|} \ln \left |\frac{\cos^2 \frac{a}{2} + \sin^2 \frac{a}{2}}{\cos^2 \frac{a}{2} - \sin^2 \frac{a}{2}} \right |\\&=\frac{a}{|\sin a|} \ln \left |\frac{1}{\cos a} \right |\\&=\frac{a}{|\sin a|} \ln |\sec a|\\&=\frac{a}{\sin a} \ln (\sec a), \quad \mbox{for}\,\, -\frac{\pi}{2} < a < \frac{\pi}{2} \end{align*}$

omegadot · Nov 27, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $ \int^\pi_0 \frac{(x + 3) \sin x}{1 + \cos^2 x} \, dx.$

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

$\\ Let\quad I=\int _{ 0 }^{ \pi }{ \arctan { \left( \cos { x } \right) } dx } \\ I=\int _{ 0 }^{ \pi }{ \arctan { \left( \cos { \left( \pi -x \right) } \right) } dx } \\ =\int _{ 0 }^{ \pi }{ \arctan { \left( -\cos { x } \right) dx } } \\ I=-\int _{ 0 }^{ \pi }{ \arctan { \left( \cos { x } \right) } dx } \\ \therefore 2I=0\Rightarrow \int _{ 0 }^{ \pi }{ \arctan { \left( \cos { x } \right) } dx } =0$

$-\int _{ 0 }^{ \pi }{ \frac { \left( x+3 \right) \left( -\sin { x } \right) }{ 1+\cos ^{ 2 }{ x } } dx } \\ =-{ \left[ \left( x+3 \right) \arctan { \left( \cos { x } \right) } \right] }_{ 0 }^{ \pi }+\int _{ 0 }^{ \pi }{ \arctan { \left( \cos { x } \right) } dx } \\ =-\left( \pi +3 \right) \arctan { \left( -1 \right) } +3\arctan { 1 } +0\\ =\frac { 3\pi }{ 2 } +\frac { { \pi }^{ 2 } }{ 4 }$

Identities used: arctan(-x)=-arctan(x); cos(pi-x)=-cos(x)
Techniques used: border flip substitution; integration by parts; reverse chain rule

Ekman · Nov 27, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Next Question}$

$$\noindent Find $ \int^\pi_0 \frac{(x + 3) \sin x}{1 + \cos^2 x} \, dx.$

ALT METHOD:

$$ Let: $ I =\int^\pi_0 \frac{(x + 3) \sin{(x)}}{1 + \cos^{2}{(x)}} \, dx = \int^\pi_0 \frac{(\pi-x + 3) \sin{(\pi-x)}}{1 + \cos^{2}{(\pi-x)}} \, dx$

$\therefore 2I = \int_{0}^{\pi} \frac{(\pi+6)\sin{(x)}}{1+\cos^{2}{(x)}} dx$

$2I = -(\pi+6) \left[\tan^{-1}{(\cos(x))} \right]_{0}^{\pi}$

$\therefore I = (\pi+6) \left(\frac{\pi}{4} \right)$

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

I should've trusted my instincts and used the border flip immediately. I forgot entirely about the classic xsin(x)/(1+cos^2(x)) cases

Drsoccerball · Nov 27, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I should've trusted my instincts and used the border flip immediately. I forgot entirely about the classic xsin(x)/(1+cos^2(x)) cases

Yeah I saw your method and said "m8 wot u doin?"

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Yeah I saw your method and said "m8 wot u doin?"

Have mercy on me it's the holidays

I'm just like ok

omegadot · Nov 27, 2015

Re: MX2 2016 Integration Marathon

A new question from anyone?

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

Someone should just do Drsoccerball's easy integral

Ekman · Nov 27, 2015

Re: MX2 2016 Integration Marathon

$\int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}(x)))) dx$

leehuan · Nov 27, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
$\int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}(x)))) dx$

Strangely familiar.

Drsoccerball · Nov 27, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
$\int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}(x)))) dx$

My final integral is : $\int{\frac{du}{u^2\sqrt{1-2u^2}}}du$

leehuan · Nov 28, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
My final integral is : $\int{\frac{du}{u^2\sqrt{1-2u^2}}}du$

RIP boundaries.

InteGrand · Nov 28, 2015

Re: MX2 2016 Integration Marathon

$Note that the integrand given in Ekman's question can be simplified to $\frac{1+x^2}{2+x^2}$, which you can show using right-angled triangles.$

Ekman · Nov 28, 2015

Re: MX2 2016 Integration Marathon

The actual answer with the boundaries are on my compilation. I cant type it out atm, so you can check it out there

kawaiipotato · Nov 28, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent The integral does not converge for all values of $ a.$ For convergence I will assume $-\frac{\pi}{2} < a < \frac{\pi}{2}.$

$\noindent Let $I = \int^a_0 \frac{x}{\cos x \cos (a - x)} \, dx.$ Using the well-known result of$

$\int^a_0 f(x) \, dx = \int^a_0 f(a - x) \, dx,$ one has$$

$\begin{align*}I &= \int^a_0 \frac{a - x}{\cos (a - x) \cos x} \, dx\\&= a \int^a_0 \frac{dx}{\cos x \cos (a - x)} - I\\\Rightarrow 2 I &= a \int^a_0 \frac{dx}{\cos x \cos (a - x)}\\\Rightarrow I &= a \int^a_0 \frac{dx}{2 \cos x \cos (a - x)}\end{align*}$

$$\noindent Applying the appropriate prosthaphaeresis formula to the denominator gives$\\\begin{align*}I &= a \int^a_0 \frac{dx}{\cos a + \cos (2x - a)}\end{align*}$

$$\noindent Let $2u = 2x - a, du = dx, u \in \left (-\frac{a}{2},\frac{a}{2} \right )$. One therefore has$

$\begin{align*}I &= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{du}{\cos a + \cos 2u}\\&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{du}{\cos a - 1 + 2 \cos^2 u}\\&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sec^2 u}{(\cos a - 1) \sec^2 u + 2} \, du&= a \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{\sec^2 u}{(\cos a + 1) + (\cos a - 1)\tan^2 u} \, du.\end{align*}$

$$\noindent Now let $t = \tan u, dt = \sec^2 u \, du, t \in \left (-\tan \frac{a}{2},\tan \frac{a}{2} \right ) = (-\alpha,\alpha)$ where $\alpha = \tan \frac{a}{2}.$ Thus$

$\begin{align*}I &= a \int^\alpha_{-\alpha} \frac{dt}{(\cos a + 1) + (\cos a - 1) t^2}\end{align*}$

$$\noindent We now note that as $0 < \cos a \leqslant 1$ for all $-\frac{\pi}{2} < a < \frac{\pi}{2}$ then $\cos a + 1 > 0$ and $\cos a - 1 \leqslant 0$. So$

$\begin{align*}I &= a \int^\alpha_{-\alpha} \frac{dt}{(\cos a + 1) - (1 - \cos a) t^2}\\&= \frac{a}{1 - \cos a} \int^\alpha_{-\alpha} \frac{dt}{\frac{1 + \cos a}{1 - \cos a} - t^2}\\&= \frac{a}{1 - \cos a} \int^{\frac{a}{2}}_{-\frac{a}{2}} \frac{dt}{\cot^2 \frac{a}{2} - t^2}, \quad \mbox{since} \,\, \cot^2 \frac{a}{2} = \frac{1 + \cos a}{1 - \cos a}\\&= \frac{a}{1 - \cos a} \cdot \frac{1}{2 \cot \frac{a}{2}} \left [\ln \left |\frac{\cot \frac{a}{2} + t}{\cot \frac{a}{2} - t} \right | \right ]^\alpha_{-\alpha}\end{align*}$

$$\noindent Here the result $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left |\frac{a + x}{a - x} \right | + \mathcal{C}$ has been used$

$\begin{align*}I &= \frac{a}{1 - \cos a} \frac{1}{2} \sqrt{\frac{1 - \cos a}{1 + \cos a}} \left [\ln \left |\frac{\cot \frac{a}{2} + \tan \frac{a}{2}}{\cot \frac{x}{2} - \tan \frac{a}{2}} \right | - \ln \left |\frac{\cot \frac{a}{2} - \tan \frac{a}{2}}{\cot \frac{x}{2} + \tan \frac{a}{2}} \right | \right ]\\&= \frac{a}{\sqrt{1 - \cos^2 a}} \ln \left |\frac{\cot \frac{a}{2} + \tan \frac{a}{2}}{\cot \frac{x}{2} - \tan \frac{a}{2}} \right |\\&= \frac{a}{|\sin a|} \ln \left |\frac{\cos^2 \frac{a}{2} + \sin^2 \frac{a}{2}}{\cos^2 \frac{a}{2} - \sin^2 \frac{a}{2}} \right |\\&=\frac{a}{|\sin a|} \ln \left |\frac{1}{\cos a} \right |\\&=\frac{a}{|\sin a|} \ln |\sec a|\\&=\frac{a}{\sin a} \ln (\sec a), \quad \mbox{for}\,\, -\frac{\pi}{2} < a < \frac{\pi}{2} \end{align*}$

Alternative method:

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

kawaiipotato said:
Alternative method:

Ah, I just knew there had to be another way.

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Here's a question for new 2016'ers I just thought of:

$\int{x^2cot^{-1}x}dx$

$$\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 + \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx = x \cot^{-1} x + \frac{1}{2} \ln (1 + x^2) + \mathcal{C}.\end{align*}$

$$\noindent Now let $I = \int x^2 \cot^{-1} \, dx. $ Applying IBP we have$\\\begin{align*}I &= x^2 \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) - \int 2x \cdot \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) \, dx\\&= x^2 \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) - 2 \int x^2 \cot^{-1} x \, dx - \int x \ln (x^2 + 1) \, dx\\\Rightarrow 3I &= x^3 \cot^{-1} x + \frac{x^2}{2} \ln (x^2 + 1) - I_1\\\Rightarrow I &= \frac{x^3}{3} \cot^{-1} x + \frac{x^2}{6} \ln (x^2 + 1) - \frac{1}{3} I_1\end{align*}$

$$\noindent For the integral $I_1$ let $u = x^2 + 1, du = 2 x \, dx$. Thus$

$\begin{align*}I_1 &= \frac{1}{2} \int \ln u \, du = \frac{1}{2} (u \ln u - u) + \mathcal{C}_1 = \frac{x^2 + 1}{2} \ln (x^2 + 1) - \frac{x^2 + 1}{2} + \mathcal{C}_1\end{align*}$

$$\noindent Thus\\\begin{align*}I &= \frac{x^3}{3} \cot^{-1} x + \frac{x^2}{6} \ln (x^2 + 1) - \frac{x^2 + 1}{6} \ln (x^2 + 1) + \frac{x^2 + 1}{6} + \mathcal{C}_1,\end{align*}$

$$\noindent or on simplifying$\\\begin{align*}\int x^2 \cot^{-1} x \, dx &= \frac{x^3}{3} \cot^{-1} x - \frac{1}{6} \ln (x^2 + 1) + \frac{x^2}{6} + \mathcal{C}.\end{align*}$

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{dx}{\cos (2x) \sin (4x)}.$

InteGrand · Nov 28, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\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 + \frac{1}{2} \int \frac{2x}{1 + x^2} \, dx = x \cot^{-1} x + \frac{1}{2} \ln (1 + x^2) + \mathcal{C}.\end{align*}$

$$\noindent Now let $I = \int x^2 \cot^{-1} \, dx. $ Applying IBP we have$\\\begin{align*}I &= x^2 \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) - \int 2x \cdot \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) \, dx\\&= x^2 \left (x \cot^{-1} x + \frac{1}{2} \ln (x^2 + 1) \right ) - 2 \int x^2 \cot^{-1} x \, dx - \int x \ln (x^2 + 1) \, dx\\\Rightarrow 3I &= x^3 \cot^{-1} x + \frac{x^2}{2} \ln (x^2 + 1) - I_1\\\Rightarrow I &= \frac{x^3}{3} \cot^{-1} x + \frac{x^2}{6} \ln (x^2 + 1) - \frac{1}{3} I_1\end{align*}$

$$\noindent For the integral $I_1$ let $u = x^2 + 1, du = 2 x \, dx$. Thus$

$\begin{align*}I_1 &= \frac{1}{2} \int \ln u \, du = \frac{1}{2} (u \ln u - u) + \mathcal{C}_1 = \frac{x^2 + 1}{2} \ln (x^2 + 1) - \frac{x^2 + 1}{2} + \mathcal{C}_1\end{align*}$

$$\noindent Thus\\\begin{align*}I &= \frac{x^3}{3} \cot^{-1} x + \frac{x^2}{6} \ln (x^2 + 1) - \frac{x^2 + 1}{6} \ln (x^2 + 1) + \frac{x^2 + 1}{6} + \mathcal{C}_1,\end{align*}$

$$\noindent or on simplifying$\\\begin{align*}\int x^2 \cot^{-1} x \, dx &= \frac{x^3}{3} \cot^{-1} x - \frac{1}{6} \ln (x^2 + 1) + \frac{x^2}{6} + \mathcal{C}.\end{align*}$

$\textbf{ALTERNATE METHOD}$

$Let $ \mathcal{I} = \int x^2 \cot ^{-1} x \text{ d}x$. Using IBP by integrating the $x^2$ and differentiating the $\cot ^{-1} x$ yields$

$\begin{align*} \mathcal{I} &= \frac{x^3}{3}\cot ^{-1}x -\frac{1}{3} \int x^3 \cdot \frac{-1}{1+x^2}\text{ d}x \\ &= \frac{x^3}{3} \cot ^{-1}x + \frac{1}{3} \int \frac{x^3}{1+x^2} \text{ d}x \\ &= \frac{x^3}{3}\cot ^{-1}x +\frac{1}{3}\int \left( x -\frac{x}{1+x^2}\right) \text{ d}x \\ &= \frac{x^3}{3}\cot^{-1}x +\frac{1}{3}\left(\frac{x^2}{2}-\frac{1}{2}\ln \left(1 + x^2\right) + \mathcal{K} \right) \\ &= \frac{x^3}{3}\cot ^{-1}x + \frac{x^2}{6} -\frac{1}{6} \ln \left( 1+x^2 \right) + \mathcal{C}\quad \Big{(}\text{where }\mathcal{C} =\frac{1}{3}\mathcal{K}\Big{)}. \end{align*}$

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