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

omegadot · Dec 1, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $ \int \frac{x^2 \tan^{-1} x}{1 + x^2} \, dx.$

Drsoccerball · Dec 1, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $ \int \frac{x^2 \tan^{-1} x}{1 + x^2} \, dx.$

IBP twice letting x^2=u and tan^(-1)x=v'

Edit: My bad tried doing it in my head.

$I= \int \frac{tan^{-1}x(1+x^2)-tan^{-1}x}{1+x^2}$

Rest is self explained.

Drsoccerball · Dec 1, 2015

Re: MX2 2016 Integration Marathon

New Question

$\int_0^{\frac{\pi}{4}}{\frac{x+sinx}{1+cosx}}dx$

omegadot · Dec 2, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
New Question

$\int_0^{\frac{\pi}{4}}{\frac{x+sinx}{1+cosx}}dx$

$\begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right ] \, dx\\&= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{2 \cos^2 \frac{x}{2}} + \tan \frac{x}{2} \right ] \, dx\\&= \frac{1}{2} \int^{\frac{\pi}{4}}_0 \left (x \sec^2 \frac{x}{2} + 2 \tan \frac{x}{2} \right ) \, dx\end{align*}$

$$\noindent Let $x = 2u, dx = 2du$. Limits: $x = 0, u = 0$ and $x = \frac{\pi}{4}, u = \frac{\pi}{8}$. Thus$

$\begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= 2 \int^{\frac{\pi}{8}}_0 \left (u \sec^2 u + \tan u \right ) \, du\\&= 2 \int^{\frac{\pi}{8}}_0 u \sec^2 u \, du + 2 \int^{\frac{\pi}{8}}_0 \tan u \, du\\&= 2 \Big{[} u \tan u \Big{]}^{\frac{\pi}{8}}_0 - 2 \int^{\frac{\pi}{8}}_0 \tan u \, du + 2 \int^{\frac{\pi}{8}}_0 \tan u \, du \qquad \mbox{(by parts)}\\&= \frac{\pi}{4} \tan \left (\frac{\pi}{8} \right ).\end{align*}$

omegadot · Dec 2, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int x \cot x \, \mbox{cosec} \, x \, dx.$

omegadot · Dec 2, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Alternatively} -- in finishing the integral off an IBP could be avoided by recognising:$

$\begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= 2 \int^{\frac{\pi}{8}}_0 \left (u \sec^2 u + \tan u \right ) \, du\\&= 2 \int^{\frac{\pi}{8}}_0 (u \tan u)' \, du\\&= \Big{[} u \tan u \Big{]}^{\frac{\pi}{8}}_0\\&= \frac{\pi}{4} \tan \left (\frac{\pi}{8} \right ),\end{align*}\\$as before.$$

Drsoccerball · Dec 2, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$\begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right ] \, dx\\&= \int^{\frac{\pi}{4}}_0 \left [\frac{x}{2 \cos^2 \frac{x}{2}} + \tan \frac{x}{2} \right ] \, dx\\&= \frac{1}{2} \int^{\frac{\pi}{4}}_0 \left (x \sec^2 \frac{x}{2} + 2 \tan \frac{x}{2} \right ) \, dx\end{align*}$

$$\noindent Let $x = 2u, dx = 2du$. Limits: $x = 0, u = 0$ and $x = \frac{\pi}{4}, u = \frac{\pi}{8}$. Thus$

$\begin{align*}\int^{\frac{\pi}{4}}_0 \frac{x + \sin x}{1 + \cos x} \, dx &= 2 \int^{\frac{\pi}{8}}_0 \left (u \sec^2 u + \tan u \right ) \, du\\&= 2 \int^{\frac{\pi}{8}}_0 u \sec^2 u \, du + 2 \int^{\frac{\pi}{8}}_0 \tan u \, du\\&= 2 \Big{[} u \tan u \Big{]}^{\frac{\pi}{8}}_0 - 2 \int^{\frac{\pi}{8}}_0 \tan u \, du + 2 \int^{\frac{\pi}{8}}_0 \tan u \, du \qquad \mbox{(by parts)}\\&= \frac{\pi}{4} \tan \left (\frac{\pi}{8} \right ).\end{align*}$

Nice but why are you making a substitution when the expression is easily integratable?
Edit: My bad didnt see your next post

porcupinetree · Dec 2, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $\int x \cot x \, \mbox{cosec} \, x \, dx.$

omegadot · Dec 3, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\ln(1 + \sin^2 x)}{\sin^2 x} dx.$

kawaiipotato · Dec 4, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $\int \frac{\ln(1 + \sin^2 x)}{\sin^2 x} dx.$

$Integrating by parts,$

$u = \ln (1+ \sin ^2 x ) . dv = cosec ^2 x dx$

$du = \frac{2 \sin x \cos x}{1 + \sin ^2 x} dx . v = - \cot x$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ \cos ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ 1- \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ - 1 + \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ 1 + \sin ^2 x - 2}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$$ Using the substitution $ t = \tan x$

$\frac{dt}{1+t^2 } = dx$

$\sin ^2 x = \frac{t^2}{1+t^2}$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int 2 - \frac{1}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4(2tan ^{-1}x - tan ^{-1}(tan ^{-1}x}) + C$
___

$$If$ f(\frac{3x-4}{3x+4}) = x + 2$

$$ Find $ \int f(x) dx$

InteGrand · Dec 4, 2015

Re: MX2 2016 Integration Marathon

kawaiipotato said:
$Integrating by parts,$

$u = \ln (1+ \sin ^2 x ) . dv = cosec ^2 x dx$

$du = \frac{2 \sin x \cos x}{1 + \sin ^2 x} dx . v = - \cot x$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ \cos ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ 1- \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ - 1 + \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ 1 + \sin ^2 x - 2}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$$ Using the substitution $ t = \tan x$

$\frac{dt}{1+t^2 } = dx$

$\sin ^2 x = \frac{t^2}{1+t^2}$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int 2 - \frac{1}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4(2tan ^{-1}x - tan ^{-1}(tan ^{-1}x}) + C$
___

$$If$ f(\frac{3x-4}{3x+4}) = x + 2$

$$ Find $ \int f(x) dx$

$\noindent Let $g(x) = \frac{3x-4}{3x+4}$, then to find $g^{-1}$, we have$

$\noindent $\begin{align*}x&=\frac{3y-4}{3y+4}\quad (\text{where }y=g^{-1}(x))\\ \Rightarrow x\cdot 3y + 4x &= 3y-4 \\ \Rightarrow 3y \left(1-x \right)&= 4 \left(x+1\right) \\ \Rightarrow y&= \frac{4 \left(x+1\right) }{3 \left(1-x \right)}. \end{align*}$$

$\noindent Now, we are given that $f\left(g(y)\right) = y+2$. Letting $y=g^{-1} \left(x\right)= \frac{4 \left(x+1\right) }{3 \left(1-x \right)} \Longleftrightarrow x = g(y)$ yields $ f\left(x \right) = \frac{4 \left(x+1\right) }{3 \left(1-x \right)}+2=-\frac{4}{3}\left(\frac{x+1}{x-1} \right)+2$. So we have$

$\noindent $\begin{align*} \int f(x)\text{ d}x &= \int \left[ -\frac{4}{3}\left(\frac{x+1}{x-1} \right)+2\right] \text{ d}x \\ &=-\frac{4}{3}\int \frac{x+1}{x-1} \text{ d}x + 2x + \mathcal{C}\\ &=-\frac{4}{3}\int \left(1 + \frac{2}{x-1} \right) \text{ d}x + 2x + \mathcal{C}\\&= -\frac{4}{3}\Big{(}x + 2 \ln \left|x-1\right| \Big{)} + 2x + \mathcal{C}\\ &= \frac{2}{3}x - \frac{8}{3} \ln \left|x-1\right| +\mathcal{C}. \end{align*}$$

InteGrand · Dec 4, 2015

Re: MX2 2016 Integration Marathon

kawaiipotato said:
$Integrating by parts,$

$u = \ln (1+ \sin ^2 x ) . dv = cosec ^2 x dx$

$du = \frac{2 \sin x \cos x}{1 + \sin ^2 x} dx . v = - \cot x$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ \cos ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ 1- \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ - 1 + \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ 1 + \sin ^2 x - 2}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$$ Using the substitution $ t = \tan x$

$\frac{dt}{1+t^2 } = dx$

$\sin ^2 x = \frac{t^2}{1+t^2}$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int 2 - \frac{1}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4(2tan ^{-1}x - tan ^{-1}(tan ^{-1}x}) + C$

$This is nice, but at the very end, your $t$ is $\tan x$, so instead of $\tan^{-1} x$'s in the final answer, some of those should be $\tan x$'s $\ddot{\smile}$.$

InteGrand · Dec 4, 2015

Re: MX2 2016 Integration Marathon

$At the end for the last integral $\Big{(}$the $\frac{1}{1+\sin^2 x}$ one$\Big{)}$, I think you accidentally substituted wrongly. It should be like this.$

$Let $\tau = \tan x, \mathrm{d}x =\frac{1}{1+\tau^2} \text{ d}\tau, \sin^2 x =\frac{\tau^2}{1+\tau^2}$. So$

$\begin{align*} 1 +\sin^2 x &= 1+\frac{\tau^2}{1+\tau^2} \\ &= \frac{1+2\tau^2}{1+\tau^2} \\ \Rightarrow \frac{1}{1+\sin^2 x} &= \frac{1+\tau^2}{1+2\tau^2} \\ \Rightarrow \frac{1}{1+\sin^2 x}\text{ d}x &= \frac{1+\tau^2 }{1+2\tau^2} \cdot\frac{1}{1+\tau^2}\text{ d}\tau \\ &= \frac{1}{1+2\tau^2}\text{ d}\tau. \end{align*}$

$So after substituting, the integral should actually be $\int \frac{1}{1+2\tau^2} \text{ d}\tau$. This will integrate to:$

$$$\begin{align*} \int \frac{1}{1+ 2\tau^2} \text{ d}\tau &= \frac{1}{\sqrt{2}}\int \frac{1}{1 + \left(\sqrt{2}\tau\right)^2 }\cdot \sqrt{2}\text{ d}\tau \\ &=\frac{1}{\sqrt{2}} \tan^{-1} \left(\sqrt{2} \tau \right) +\mathcal{C} \\&=\frac{1}{\sqrt{2}} \tan^{-1}\left(\sqrt{2}\tan x\right)+\mathcal{C}. \end{align*}$

InteGrand · Dec 4, 2015

Re: MX2 2016 Integration Marathon

kawaiipotato said:
$Integrating by parts,$

$u = \ln (1+ \sin ^2 x ) . dv = cosec ^2 x dx$

$du = \frac{2 \sin x \cos x}{1 + \sin ^2 x} dx . v = - \cot x$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ \cos ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ 1- \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ - 1 + \sin ^2 x}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2 \int \frac{ 1 + \sin ^2 x - 2}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1}{1+ \sin ^2 x} dx$

$$ Using the substitution $ t = \tan x$

$\frac{dt}{1+t^2 } = dx$

$\sin ^2 x = \frac{t^2}{1+t^2}$

$$I$ = - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dx$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int \frac{1+2t^2}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4 \int 2 - \frac{1}{1+t^2} dt$

$= - \cot x \ln (1+ \sin ^2 x ) - 2x + 4(2tan ^{-1}x - tan ^{-1}(tan ^{-1}x}) + C$
___

$$If$ f(\frac{3x-4}{3x+4}) = x + 2$

$$ Find $ \int f(x) dx$

$By the way, an easier way to integrate the $\frac{1}{1+\sin ^2 x}$ is to divide top and bottom by $\cos^2 x$ to get $\frac{\sec^2 x}{\sec^2 x + \tan^2 x} = \frac{\sec ^2 x}{1+2\tan^2 x}$, which results in the same integral as before upon substituting $\tau =\tan x, \frac{\mathrm{d}\tau}{\mathrm{d}x} =\sec^2 x$.$

omegadot · Dec 4, 2015

Re: MX2 2016 Integration Marathon

$$\noindent And now for some integration `humour.'$

$$\noindent Find $\int \frac{dx}{dx}.$

InteGrand · Dec 4, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent And now for some integration `humour.'$

$$\noindent Find $\int \frac{dx}{dx}.$

$$\noindent Well, I suppose the ``humorous'' answer would be $\frac{1}{d}\cdot \ln \left|x\right|+C$.$

omegadot · Dec 4, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$$\noindent Well, I suppose the ``humorous'' answer would be $\frac{1}{d}\cdot \ln \left|x\right|+C$.$

Correct!

leehuan · Dec 4, 2015

Re: MX2 2016 Integration Marathon

Aiya...
--------
$\\ \int _{ 0 }^{ 1 }{ \ln { \left( \sqrt { 1+x } +\sqrt { 1-x } \right) dx } }$

omegadot · Dec 4, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Aiya...
--------
$\\ \int _{ 0 }^{ 1 }{ \ln { \left( \sqrt { 1+x } +\sqrt { 1-x } \right) dx } }$

$$\noindent We will use IBP. We begin by noting that$

$\begin{align*}\frac{d}{dx} \left [\ln \left (\sqrt{1 + x} + \sqrt{1 - x} \right ) \right ] &= \frac{2 \sqrt{1 - x} - 2 \sqrt{1 + x}}{4 \sqrt{1 - x} \sqrt{1 + x}} \cdot \frac{1}{\sqrt{1 + x} + \sqrt{1 - x}}\\&=\frac{1}{2 \sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - x} - \sqrt{1 + x}}{\sqrt{1 + x} + \sqrt{1 - x}} \cdot \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}}\\&= \frac{1}{2 \sqrt{1 - x^2}} \cdot \frac{2 \sqrt{1 - x^2} - 2}{2x}\\&= \frac{1}{2x} - \frac{1}{2x \sqrt{1 - x^2}}\end{align*}$

$\begin{align*} \int^1_0 \ln \left( \sqrt {1+x} +\sqrt {1-x} \right) \, dx &= \left [x \ln \left (\sqrt {1+x} +\sqrt {1-x} \right) \right ]^1_0 - \int^1_0 x \cdot \left (\frac{1}{2x} - \frac{1}{2x \sqrt{1 - x^2}} \right ) \, dx\\&= \ln (\sqrt{2}) - \frac{1}{2} \int^1_0 dx + \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{1 - x^2}}\\&= \ln \sqrt{2} - \frac{1}{2} \Big{[}x \Big{]}^1_0 + \frac{1}{2} \Big{[} \sin^{-1} x \Big{]}^1_0\\&= \ln (\sqrt{2}) - \frac{1}{2} + \frac{\pi}{4}\end{align*}$

Paradoxica · Dec 4, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Aiya...
--------
$\\ \int _{ 0 }^{ 1 }{ \ln { \left( \sqrt { 1+x } +\sqrt { 1-x } \right) dx } }$

$= \frac{1}{2} \int_0^1 \ln \left ( 1 - x + 1 + x + 2\sqrt{1-x^2} \right ) dx = \frac{1}{2} \int \left ( \ln 2 + \ln \left ( 1 + \sqrt{1-x^2} \right ) \right ) dx$

$= \frac{\ln 2}{2} + \frac{1}{2}\int_0^1 \ln \left ( 1 + \sqrt{1-x^2} \right ) dx = \frac{\ln 2}{2} + \frac{x}{2} \ln \left ( 1 + \sqrt{1-x^2} \right )_0^1 - \frac{1}{2} \int_0^1 x \frac{-\frac{x}{\sqrt{1-x^2}}}{1 + \sqrt{1-x^2}} dx$

$= \frac{\ln 2}{2} + \frac{1}{2} \int_0^1 \frac{x^2}{(1+ \sqrt{1-x^2})\sqrt{1-x^2}} dx = \frac{\ln 2}{2} + \frac{1}{2} \int_0^1 \frac{1^2 - (1-x^2)}{(1+ \sqrt{1-x^2})\sqrt{1-x^2}} dx$

$= \frac{\ln 2}{2} + \frac{1}{2} \int_0^1 \frac{1 - \sqrt{1-x^2}}{\sqrt{1-x^2}} dx = \frac{\ln 2}{2} + \frac{1}{2} \left ( \sin^{-1}{x} - x \right )_0^1 = \frac{\ln 2}{2} + \frac{\pi}{4} - \frac{1}{2}$

$I = \frac{2\ln 2 + \pi -2}{4}$

$$\noindent \textbf{Next Question}$

$(a,b) \in \mathbb R^+$

$$Evaluate $ \int_{-\infty}^\infty \frac{dx}{a^x + b^{\frac{1}{x}}}$

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