HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

Paradoxica · Nov 11, 2015

Re: MX2 2016 Integration Marathon

Continuation of the above
$$Evaluate $\int_0^1 \frac{x^8 (1-x)^8 (25+816x^2)}{3164(1+x^2)}dx $ and hence show that $\frac{355}{113} > \pi$

leehuan · Nov 11, 2015

Re: MX2 2016 Integration Marathon

glittergal96 said:
Your integral should be

$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\, dx.$

Will leave this for a current student to do though.

Right, my bad. I'll fix that up

omegadot · Nov 14, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
If any 2016'ers are brave enough...

$\int \sqrt{tanx}\ dx$

$No need to scare the children away so soon!$

$$\noindent We begin by noting that$\\\begin{align*}\sqrt{\tan x} &= \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\cot x} - \frac{1}{2} \sqrt{\cot x}\\&= \frac{1}{2} \left (\sqrt{\tan x} + \sqrt{\cot x} \right ) + \frac{1}{2} \left (\sqrt{\tan x} - \sqrt{\cot x} \right )\\&=\frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right ) + \frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} - \sqrt{\frac{\cos x}{\sin x}} \right )\\&=\frac{1}{2} \left (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \right ) + \frac{1}{2} \left (\frac{\sin x - \cos x}{\sqrt{\sin x \cos x}} \right )\end{align*}$

$$\noindent Now noting that$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin^2 x + \cos^2 x) + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin x - \cos x)^2}\end{align*}$

$$\noindent Similarly$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin^2 x + \cos^2 x) -1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin x + \cos x)^2 - 1}\end{align*}$

$$\noindent So for the integral we have$\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx - \frac{1}{\sqrt{2}} \int \frac{\cos x - \sin x}{\sqrt{(\cos x + \sin x)^2}} \, dx\end{align*}$

$$\noindent If we let $u = \sin x - \cos x$ in the first integral and $v = \sin x + \cos x$ in the second we have\\\begin{align*}du = (\sin x + \cos x) dx \quad &\mbox{and} \quad dv = (\cos x - \sin x) dx\end{align*}$

$$\noindent We have\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{1 - u^2}} - \frac{1}{\sqrt{2}} \int \frac{dv}{\sqrt{v^2 - 1}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} u - \frac{1}{\sqrt{2}} \ln \left |v + \sqrt{v^2 - 1} \right | + {\cal{C}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} (\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln \left |\sin x + \cos x + \sqrt{\sin 2x} \right | + {\cal{C}}\end{align*}$

porcupinetree · Nov 14, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$No need to scare the children away so soon!$

$$\noindent We begin by noting that$\\\begin{align*}\sqrt{\tan x} &= \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\cot x} - \frac{1}{2} \sqrt{\cot x}\\&= \frac{1}{2} \left (\sqrt{\tan x} + \sqrt{\cot x} \right ) + \frac{1}{2} \left (\sqrt{\tan x} - \sqrt{\cot x} \right )\\&=\frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right ) + \frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} - \sqrt{\frac{\cos x}{\sin x}} \right )\\&=\frac{1}{2} \left (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \right ) + \frac{1}{2} \left (\frac{\sin x - \cos x}{\sqrt{\sin x \cos x}} \right )\end{align*}$

$$\noindent Now noting that$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin^2 x + \cos^2 x) + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin x - \cos x)^2}\end{align*}$

$$\noindent Similarly$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin^2 x + \cos^2 x) -1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin x + \cos x)^2 - 1}\end{align*}$

$$\noindent So for the integral we have$\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx - \frac{1}{\sqrt{2}} \int \frac{\cos x - \sin x}{\sqrt{(\cos x + \sin x)^2}} \, dx\end{align*}$

$$\noindent If we let $u = \sin x - \cos x$ in the first integral and $v = \sin x + \cos x$ in the second we have\\\begin{align*}du = (\sin x + \cos x) dx \quad &\mbox{and} \quad dv = (\cos x - \sin x) dx\end{align*}$

$$\noindent We have\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{1 - u^2}} - \frac{1}{\sqrt{2}} \int \frac{dv}{\sqrt{v^2 - 1}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} u - \frac{1}{\sqrt{2}} \ln \left |v + \sqrt{v^2 - 1} \right | + {\cal{C}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} (\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln \left |\sin x + \cos x + \sqrt{\sin 2x} \right | + {\cal{C}}\end{align*}$

Never seen this integral done this way before, good job

leehuan · Nov 14, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$No need to scare the children away so soon!$

$$\noindent We begin by noting that$\\\begin{align*}\sqrt{\tan x} &= \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\tan x} + \frac{1}{2} \sqrt{\cot x} - \frac{1}{2} \sqrt{\cot x}\\&= \frac{1}{2} \left (\sqrt{\tan x} + \sqrt{\cot x} \right ) + \frac{1}{2} \left (\sqrt{\tan x} - \sqrt{\cot x} \right )\\&=\frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right ) + \frac{1}{2} \left (\sqrt{\frac{\sin x}{\cos x}} - \sqrt{\frac{\cos x}{\sin x}} \right )\\&=\frac{1}{2} \left (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \right ) + \frac{1}{2} \left (\frac{\sin x - \cos x}{\sqrt{\sin x \cos x}} \right )\end{align*}$

$$\noindent Now noting that$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin^2 x + \cos^2 x) + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{1 - (\sin x - \cos x)^2}\end{align*}$

$$\noindent Similarly$\\\begin{align*}\sqrt{\sin x \cos x} &= \frac{1}{\sqrt{2}} \sqrt{2 \sin x \cos x}\\&= \frac{1}{\sqrt{2}} \sqrt{1 - 1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin^2 x + \cos^2 x) -1 + 2 \sin x \cos x}\\&=\frac{1}{\sqrt{2}} \sqrt{(\sin x + \cos x)^2 - 1}\end{align*}$

$$\noindent So for the integral we have$\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx - \frac{1}{\sqrt{2}} \int \frac{\cos x - \sin x}{\sqrt{(\cos x + \sin x)^2}} \, dx\end{align*}$

$$\noindent If we let $u = \sin x - \cos x$ in the first integral and $v = \sin x + \cos x$ in the second we have\\\begin{align*}du = (\sin x + \cos x) dx \quad &\mbox{and} \quad dv = (\cos x - \sin x) dx\end{align*}$

$$\noindent We have\\\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \int \frac{du}{\sqrt{1 - u^2}} - \frac{1}{\sqrt{2}} \int \frac{dv}{\sqrt{v^2 - 1}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} u - \frac{1}{\sqrt{2}} \ln \left |v + \sqrt{v^2 - 1} \right | + {\cal{C}}\\&= \frac{1}{\sqrt{2}} \sin^{-1} (\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln \left |\sin x + \cos x + \sqrt{\sin 2x} \right | + {\cal{C}}\end{align*}$

Back when I did it I just bombarded it with u^2=tan(x) and worked from there. From memory it requires a second substitution as well as partial fractions that way.

But yeah, blame the porcupine for scaring away the young ones

Paradoxica · Nov 15, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Back when I did it I just bombarded it with u^2=tan(x) and worked from there. From memory it requires a second substitution as well as partial fractions that way.

But yeah, blame the porcupine for scaring away the young ones

That add and subtract method I have seen before somewhere on math.stackexchange...

Anyway, an easier one for the current students.

$$The formula for the arc length of a simple plane curve given as $y = R(x)$ is given by $\int_a^b \sqrt{1+\left ( \frac{dy}{dx} \right )^2}dx$

$$Using this formula on the unit circle, deduce that the arc length of the curved portion of a unit semicircle is $\pi$ units.$

omegadot · Nov 15, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
Never seen this integral done this way before, good job

$\noindent Or for a third approach which also avoids the dreaded partial fractions, start with the standard substitution of $u^2 = \tan x$ to arrive at$

$\int \sqrt{\tan x} \, dx = \int \frac{2u^2}{1+u^4} du.$

$$\noindent Now rather than launching head long into partial fractions, considering manipulating the integrand as follows:$\\\begin{align*}\int \sqrt{\tan x} \, dx &= \int \frac{u^2 + 1 + u^2 - 1}{1 + u^4} du\\&= \int \frac{u^2 + 1}{1+ u^4} du + \int \frac{u^2 - 1}{1 + u^4} du\\&= \int \frac{u^2 + 1}{u^2(1/u^2 + u^2)} du + \int \frac{u^2 - 1}{u^2 (1/u^2 + u^2)} du\\&=\int \frac{1 + \frac{1}{u^2}}{\frac{1}{u^2} + u^2} du + \int \frac{1 - \frac{1}{u^2}}{\frac{1}{u^2} + u^2} du\\&=\int \frac{1 + \frac{1}{u^2}}{\left (u - \frac{1}{u} \right )^2 + 2} du + \int \frac{1 - \frac{1}{u^2}}{\left (u + \frac{1}{u} \right )^2 - 2}\end{align*}$

$$\noindent In the first integral we make the substitution $t = u - \frac{1}{u}$ while in the second $v = u + \frac{1}{u}$ which gives:\\\begin{align*}dt = \left (1 + \frac{1}{u^2} \right ) du \quad \mbox{and} \quad dv = \left (1 - \frac{1}{u^2} \right ) du.\end{align*}$

$$\noindent So the two integrals become\\\begin{align*}\int \sqrt{\tan x} \, dx &= \int \frac{dt}{t^2 + 2} + \int \frac{dv}{v^2 - 2}\\&= \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{t}{\sqrt{2}} \right ) + \frac{1}{2\sqrt{2}} \ln \left |\frac{v - \sqrt{2}}{v + \sqrt{2}} \right | + {\cal{C}}\\&= \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{1}{\sqrt{2}} \left (u - \frac{1}{u} \right ) \right ) + \frac{1}{2\sqrt{2}} \ln \left |\frac{u + \frac{1}{u} - \sqrt{2}}{u + \frac{1}{u} + \sqrt{2}} \right | + {\cal{C}}\\&= \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{\tan x - 1}{\sqrt{2 \tan x}} \right ) + \frac{1}{2\sqrt{2}} \ln \left |\frac{\tan x - \sqrt{2 \tan x} + 1}{\tan x + \sqrt{2 \tan x} + 1} \right | + \cal{C}\end{align*}$

KingOfActing · Nov 15, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
That add and subtract method I have seen before somewhere on math.stackexchange...

Anyway, an easier one for the current students.

$$The formula for the arc length of a simple plane curve given as $y = R(x)$ is given by $\int_a^b \sqrt{1+\left ( \frac{dy}{dx} \right )^2}dx$

$$Using this formula on the unit circle, deduce that the arc length of the curved portion of a unit semicircle is $\pi$ units.$

$\begin{align*} &\text{Unit circle: } x^2 + y^2 = 1 \\ &2x\ dx + 2y\ dy = 0 \Longleftrightarrow \frac{dy}{dx} = -\frac{x}{y} \\&\text{Using the arc length formula, the arc length of the semicircle is} \int_{-1}^{1}\sqrt{1 + \left( -\frac{x}{y}\right)^2}\ dx\\&= \int_{-1}^{1}\sqrt{1 + \frac{x^2}{y^2}}\ dx = \int_{-1}^{1}\sqrt{1 + \frac{x^2}{1-x^2}}\ dx = \int_{-1}^{1}\sqrt{\frac{1}{1-x^2}}\ dx\\&\text{Let } x = \sin{\theta}\\&\text{Our integral becomes } \int_{x=-1}^{1}\cos{\theta}\sec{\theta}\ d\theta = \int_{x=-1}^{1}d\theta = \theta\Big|_{x=-1}^{1} = \arcsin{x}\Big|_{-1}^{1} \\&=\arcsin{(1)} - \arcsin{(-1)} = \frac{\pi}{2} - \frac{-\pi}{2} = \pi \end{align*}$

Sy123 · Nov 15, 2015

Re: MX2 2016 Integration Marathon

$\\ $i) Prove that$ \ \int_0^a f(x) \ dx = \int_0^a f(a-x) \ dx$

$\\ $ii) Hence find$ \ \int_0^{\pi} \frac{x \sin x}{1 + \sin^2x } \ dx$

KingOfActing · Nov 15, 2015

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ $i) Prove that$ \ \int_0^a f(x) \ dx = \int_0^a f(a-x) \ dx$

$\\ $ii) Hence find$ \ \int_0^{\pi} \frac{x \sin x}{1 + \sin^2x } \ dx$

$\begin{align*}&\text{i) }\int_{0}^{a}f(x)\ dx = F(a) - F(0) \\&\int_{0}^{a}f(a-x)\ dx = -\int_{u=a}^{0}f(u)\ du\;\; \text{Where } u = a-x\\&=-(F(0)-F(a)) = F(a) - F(0)\end{align*}\\\begin{align*}\\\text{ii)} \int_{0}^{\pi}\frac{x\sin{x}}{1+\sin^2{x}}\ dx &= \int_{0}^{\pi} \frac{(\pi-x)\sin{(\pi-x)}}{1+\sin^2{(\pi-x)}}\ dx\\&=\int_{0}^{\pi}\frac{(\pi-x)\sin{x}}{1+\sin^2{x}}\ dx\\&=\pi\int_{0}^{\pi}\frac{\sin{x}}{1+\sin^2{x}}\ dx - \int_{0}^{\pi}\frac{x\sin{x}}{1+\sin^2{x}}\ dx\end{align*}\\$By letting the original integral $ = I $ we obtain $ \\\begin{align*} I &= \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin{x}}{1+\sin^2{x}} \ dx\end{align*}$

That's as far as I could get, I'm not sure how to evaluate this integral.

InteGrand · Nov 15, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\begin{align*}&\text{i) }\int_{0}^{a}f(x)\ dx = F(a) - F(0) \\&\int_{0}^{a}f(a-x)\ dx = -\int_{u=a}^{0}f(u)\ du\;\; \text{Where } u = a-x\\&=-(F(0)-F(a)) = F(a) - F(0)\end{align*}\\\begin{align*}\\\text{ii)} \int_{0}^{\pi}\frac{x\sin{x}}{1+\sin^2{x}}\ dx &= \int_{0}^{\pi} \frac{(\pi-x)\sin{(\pi-x)}}{1+\sin^2{(\pi-x)}}\ dx\\&=\int_{0}^{\pi}\frac{(\pi-x)\sin{x}}{1+\sin^2{x}}\ dx\\&=\pi\int_{0}^{\pi}\frac{\sin{x}}{1+\sin^2{x}}\ dx - \int_{0}^{\pi}\frac{x\sin{x}}{1+\sin^2{x}}\ dx\end{align*}\\$By letting the original integral $ = I $ we obtain $ \\\begin{align*} I &= \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin{x}}{1+\sin^2{x}} \ dx\end{align*}$

That's as far as I could get, I'm not sure how to evaluate this integral.

$For the last integral, we can replace $\sin^2 x$ with $1-\cos^2 x$ and then substitute $u=\cos x$, $\mathrm{d}u = -\sin x \text{ d}x$.$

leehuan · Nov 16, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\begin{align*}&\text{i) }\int_{0}^{a}f(x)\ dx = F(a) - F(0) \\&\int_{0}^{a}f(a-x)\ dx = -\int_{u=a}^{0}f(u)\ du\;\; \text{Where } u = a-x\\&=-(F(0)-F(a)) = F(a) - F(0)\end{align*}$

There is a slightly more elegant approach to this than evaluating both integrals. After using the substitution u=a-x, the integral becomes $\int_{0}^{a}f(u)\ du$ , but because this is a definite integral we can just use dummy variables and say that's equal to $\int_{0}^{a}f(x)\ dx$

Note that negative signs can be removed by interchanging the lower and upper bounds.

Sy123 · Nov 16, 2015

Re: MX2 2016 Integration Marathon

$\\ $Find$ \ \int_{-\pi}^{\pi} \sin(nx) \sin(mx) \ dx \ $in the cases where$ \ m \neq n \ $and$ \ m = n \ $and$ \ m,n \ $are integers$$

omegadot · Nov 16, 2015

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ $Find$ \ \int_{-\pi}^{\pi} \sin(nx) \sin(mx) \ dx \ $in the cases where$ \ m \neq n \ $and$ \ m = n \ $and$ \ m,n \ $are integers$$

$$\noindent For all integers, in which case $m = n = 0$ will also need to be considered separately, or for just positive integers $ m$ and $n$ only?$

leehuan · Nov 17, 2015

Re: MX2 2016 Integration Marathon

$\\ $In the case where $m \neq\ n, m \neq 0, n \neq 0$:$ \\ \cos { \left( A-B \right) } -\cos { \left( A+B \right) } =\left( \cos { A } \cos { B } +\sin { A } \sin { B } \right) -\left( \cos { A } \cos { B } -\sin { A } \sin { B } \right) \\ \frac { 1 }{ 2 } \left( \cos { \left( A-B \right) } -\cos { \left( A+B \right) } \right) =\sin { A } \sin { B } \\ \int _{ -\pi }^{ \pi }{ \sin { \left( nx \right) } \sin { \left( mx \right) } dx } \\ =\int _{ -\pi }^{ \pi }{ \left( \cos { \left[ \left( m-n \right) x \right] } -\cos { \left[ \left( m+n \right) x \right] } \right) dx } \\ ={ \left[ \frac { \sin { \left[ \left( m-n \right) x \right] } }{ m-n } -\frac { \sin { \left[ \left( m+n \right) x \right] } }{ m+n } \right] }_{ -\pi }^{ \pi }\\ =0\\ \because \sin { \left( k\pi \right) } =0\forall k\in \mathbb Z$

$\\ $In the case where $m=n \neq 0 $:$ \\ \int _{ -\pi }^{ \pi }{ \sin { \left( mx \right) } \sin { \left( nx \right) } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ \left( nx \right) } dx } \\ =2\int _{ 0 }^{ \pi }{ \sin ^{ 2 }{ \left( nx \right) } dx } \\ \because f\left( x \right) =\sin ^{ 2 }{ \left( nx \right) } $ is even$\\ =\int _{ 0 }^{ \pi }{ \left( 1-\cos { \left( 2nx \right) } \right) dx } \\ ={ \left[ x-\frac { \sin { \left( 2nx \right) } }{ 2n } \right] }_{ 0 }^{ \pi }\\ =\left( \pi -0 \right) -\left( 0-0 \right) \\ =\pi$

In the special case when m=0 and/or n=0, we are integrating the zero function as sin(0)=0, so the final answer will trivially be 0.
------------------------------
NEXT QUESTION
$\int { \arctan { x } \, dx }$

KingOfActing · Nov 17, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
NEXT QUESTION
$\int { \arctan { x } \, dx }$

$\begin{align*} \int\arctan{x}\,dx &= \int\theta\sec^2{\theta}\,d\theta \;\;\left( \text{By letting } x = \tan{\theta}\right) \\&= \theta\tan{\theta} - \int{\tan{\theta}\,d\theta} \;\; \left(\text{Integration by parts} \right ) \\&=\theta\tan{\theta} + \ln{\cos{\theta}} + C \\&= x\arctan{x} + \ln{\frac{1}{\sqrt{x^2+1}}} + C \\&=x\arctan{x} - \frac{1}{2}\ln{(x^2+1)} + C\end{align*}$

leehuan · Nov 17, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\begin{align*} \int\arctan{x}\,dx &= \int\theta\sec^2{\theta}\,d\theta \;\;\left( \text{By letting } x = \tan{\theta}\right) \\&= \theta\tan{\theta} - \int{\tan{\theta}\,d\theta} \;\; \left(\text{Integration by parts} \right ) \\&=\theta\tan{\theta} + \ln{\cos{\theta}} + C \\&= x\arctan{x} + \ln{\frac{1}{\sqrt{x^2+1}}} + C \\&=x\arctan{x} - \frac{1}{2}\ln{(x^2+1)} + C\end{align*}$

Correct, although you could've done IBP immediately by differentiating arctan(x) and integrating 1.

KingOfActing · Nov 17, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Correct, although you could've done IBP immediately by differentiating arctan(x) and integrating 1.

I didn't think of that o:

leehuan · Nov 17, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
I didn't think of that o:

Classic trick to integrating the inverse trig functions. Put it in your toolbox!

KingOfActing · Nov 17, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Classic trick to integrating the inverse trig functions. Put it in your toolbox!

I will definitely remember that, thanks!

NEW QUESTION:
$\int\sin{(\ln{x})}\,dx$

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