HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

FrankXie · May 26, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Find $\int \frac{x^2}{1+x^4}\text{ d}x$.$

split into partial fractions $\frac{x^2}{1+x^4} = \frac{1}{2\sqrt2}\left(\frac{x}{1+x^2-\sqrt2x}-\frac{x}{1+x^2+\sqrt2x}\right)$

leehuan · May 27, 2015

Re: MX2 2015 Integration Marathon

$\int { \frac { x-1 }{ { x }^{ 6 }-1 } dx }$

kawaiipotato · May 27, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { x-1 }{ { x }^{ 6 }-1 } dx }$

factorise (x^6 - 1) and you'll eventually get int (dx/(x^2 + x+1)(x^3 + 1)) and then partial
Idk if there's an easier way

braintic · May 27, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { x-1 }{ { x }^{ 6 }-1 } dx }$

Find the 6th roots of unity, delete 1 from the list, pair the complex roots to form quadratic factors, then use partial fractions.
(or use partial fractions on the complex factors)

leehuan · May 28, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
factorise (x^6 - 1) and you'll eventually get int (dx/(x^2 + x+1)(x^3 + 1)) and then partial
Idk if there's an easier way

Just remember to break the (x^3+1) further into (x+1)(x^2-x+1) before you commence the partial fractions process.

What I would've done:
${ x }^{ 6 }-1\\ =\left( x-1 \right) \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1 \right) \\ But\\ { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1\\ ={ x }^{ 4 }\left( x+1 \right) +{ x }^{ 2 }\left( x+1 \right) +\left( x+1 \right) \\ =\left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) \left( x+1 \right) \\ =\left( \left( { x }^{ 4 }+2{ x }^{ 2 }+1 \right) -{ x }^{ 2 } \right) \left( x+1 \right) \\ =\left( { \left( { x }^{ 2 }+1 \right) }^{ 2 }-{ x }^{ 2 } \right) \left( x+1 \right) \\ =\left( { x }^{ 2 }+1+x \right) \left( { x }^{ 2 }+1-x \right) \left( x+1 \right)$

When I made the problem I forgot you could factor out (x^2-1) because it was a difference of even powers.

Ekman · May 28, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \frac{\sec{x}}{2+\sin^2{x}} dx$

leehuan · May 28, 2015

Re: MX2 2015 Integration Marathon

$u=\sin { x } \Rightarrow du=\cos { x } dx\\ \int { \frac { \sec { x } }{ 2+\sin ^{ 2 }{ x } } dx } \\ =\int { \frac { \cos { x } }{ 2\cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx\\ =\int { \frac { \cos { x } }{ 2\left( 1-\sin ^{ 2 }{ x } \right) +\sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) } } dx\\ =\int { \frac { du }{ 2\left( 1-{ u }^{ 2 } \right) +{ u }^{ 2 }\left( 1-{ u }^{ 2 } \right) } } \\ =\int { \frac { du }{ \left( 2+{ u }^{ 2 } \right) \left( 1+u \right) \left( 1-u \right) } } \\ Partial\quad fractions...$
$=\int { \left( \frac { 1 }{ 3\left( { u }^{ 2 }+2 \right) } +\frac { 1 }{ 6\left( u+1 \right) } -\frac { 1 }{ 6\left( u-1 \right) } \right) du } \\ =\frac { 1 }{ 3\sqrt { 2 } } \tan ^{ -1 }{ \frac { u }{ \sqrt { 2 } } } +\frac { 1 }{ 6 } \ln { \left| u+1 \right| } -\frac { 1 }{ 6 } \ln { \left| u-1 \right| } +C\\ =\frac { 1 }{ 3\sqrt { 2 } } \tan ^{ -1 }{ \frac { \sin { x } }{ \sqrt { 2 } } } +\frac { 1 }{ 6 } \ln { \left| \frac { \sin { x } +1 }{ \sin { x } -1 } \right| } +C$

Ekman · May 28, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$u=\sin { x } \Rightarrow du=\cos { x } dx\\ \int { \frac { \sec { x } }{ 2+\sin ^{ 2 }{ x } } dx } \\ =\int { \frac { \cos { x } }{ 2\cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx\\ =\int { \frac { \cos { x } }{ 2\left( 1-\sin ^{ 2 }{ x } \right) +\sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) } } dx\\ =\int { \frac { du }{ 2\left( 1-{ u }^{ 2 } \right) +{ u }^{ 2 }\left( 1-{ u }^{ 2 } \right) } } \\ =\int { \frac { du }{ \left( 2+{ u }^{ 2 } \right) \left( 1+u \right) \left( 1-u \right) } } \\ Partial\quad fractions...$
$=\int { \left( \frac { 1 }{ 3\left( { u }^{ 2 }+2 \right) } +\frac { 1 }{ 6\left( u+1 \right) } -\frac { 1 }{ 6\left( u-1 \right) } \right) du } \\ =\frac { 1 }{ 3\sqrt { 2 } } \tan ^{ -1 }{ \frac { u }{ \sqrt { 2 } } } +\frac { 1 }{ 6 } \ln { \left| u+1 \right| } -\frac { 1 }{ 6 } \ln { \left| u-1 \right| } +C\\ =\frac { 1 }{ 3\sqrt { 2 } } \tan ^{ -1 }{ \frac { \sin { x } }{ \sqrt { 2 } } } +\frac { 1 }{ 6 } \ln { \left| \frac { \sin { x } +1 }{ \sin { x } -1 } \right| } +C$

Just to shorten the first bit of working out:

$\int \frac{\sec{x}}{2+\sin^2{x}}dx =\int \frac{1}{\cos{x}(2+\sin^2{x})}dx$

So when you change the dx to du for the substitution u=sinx, you will have:

$\int \frac{1}{\cos^{2}{x}(2+u^2)}du$

So using the property sin^2x +cos^2x =1, you can directly go to the 4th step.

Other than that it looks all good.

Ekman · May 28, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \frac{\sin{x}}{\sqrt{\sin{x}+1}} dx$

InteGrand · May 28, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \frac{\sin{x}}{\sqrt{\sin{x}+1}} dx$

$Let $I= \int \frac{\sin{x}}{\sqrt{\sin{x}+1}} \text{ d}x$.$

$I=\int \frac{\sin{x}+1}{\sqrt{\sin{x}+1}} \text{ d}x - \int \frac{1}{\sqrt{\sin{x}+1}} \text{ d}x$

$=\int \sqrt{\sin{x}+1} \text{ d}x - \int \frac{1}{\sqrt{\sin{x}+1}} \text{ d}x.$

$Note that $\sin x + 1 = 2\sin \frac{x}{2}\cos \frac{x}{2} + \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$$

$=\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2$

$\Rightarrow \sqrt{\sin{x}+1}=\sin \frac{x}{2} + \cos \frac{x}{2}.$

$Hence $I=\int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \text{ d}x - \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} \text{ d}x$$

$$=\mathcal{J}_1 - \mathcal{J}_2$, where $\mathcal{J}_1 = \int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \text{ d}x$ and $\mathcal{J}_2 = \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} \text{ d}x$.$

$Clearly $\mathcal{J}_1 = 2\left(\sin \frac{x}{2} -\cos \frac{x}{2} \right)$. To find $\mathcal{J}_2$, we note that $\sin \frac{x}{2} + \cos \frac{x}{2} \equiv \sqrt{2} \sin \left(\frac{x}{2}+\frac{\pi}{4} \right)$, so$

$\mathcal{J}_2 = \int \frac{1}{\sqrt{2} \sin \left(\frac{x}{2}+\frac{\pi}{4} \right)} \text{ d}x$

$= \frac{\sqrt{2}}{2}\int \csc \left(\frac{x}{2}+\frac{\pi}{4} \right)} \text{ d}x$

$= -\sqrt{2}\ln \left|\csc \left(\frac{x}{2}+\frac{\pi}{4} \right) + \cot \left(\frac{x}{2}+\frac{\pi}{4} \right) \right|$

$\Rightarrow I = \mathcal{J}_1 - \mathcal{J}_2$

$=2\left(\sin \frac{x}{2} -\cos \frac{x}{2} \right) + \sqrt{2}\ln \left|\csc \left(\frac{x}{2}+\frac{\pi}{4} \right) + \cot \left(\frac{x}{2}+\frac{\pi}{4} \right) \right| +c.$

Ekman · May 28, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $I= \int \frac{\sin{x}}{\sqrt{\sin{x}+1}} \text{ d}x$.$

$I=\int \frac{\sin{x}+1}{\sqrt{\sin{x}+1}} \text{ d}x - \int \frac{1}{\sqrt{\sin{x}+1}} \text{ d}x$

$=\int \sqrt{\sin{x}+1} \text{ d}x - \int \frac{1}{\sqrt{\sin{x}+1}} \text{ d}x.$

$Note that $\sin x + 1 = 2\sin \frac{x}{2}\cos \frac{x}{2} + \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$$

$=\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2$

$\Rightarrow \sqrt{\sin{x}+1}=\sin \frac{x}{2} + \cos \frac{x}{2}.$

$Hence $I=\int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \text{ d}x - \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} \text{ d}x$$

$$=\mathcal{J}_1 - \mathcal{J}_2$, where $\mathcal{J}_1 = \int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \text{ d}x$ and $\mathcal{J}_2 = \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} \text{ d}x$.$

$Clearly $\mathcal{J}_1 = 2\left(\sin \frac{x}{2} -\cos \frac{x}{2} \right)$. To find $\mathcal{J}_2$, we note that $\sin \frac{x}{2} + \cos \frac{x}{2} \equiv \sqrt{2} \sin \left(\frac{x}{2}+\frac{\pi}{4} \right)$, so$

$\mathcal{J}_2 = \int \frac{1}{\sqrt{2} \sin \left(\frac{x}{2}+\frac{\pi}{4} \right)} \text{ d}x$

$= \frac{\sqrt{2}}{2}\int \csc \left(\frac{x}{2}+\frac{\pi}{4} \right)} \text{ d}x$

$= -\sqrt{2}\ln \left|\csc \left(\frac{x}{2}+\frac{\pi}{4} \right) + \cot \left(\frac{x}{2}+\frac{\pi}{4} \right) \right|$

$\Rightarrow I = \mathcal{J}_1 - \mathcal{J}_2$

$=2\left(\sin \frac{x}{2} -\cos \frac{x}{2} \right) + \sqrt{2}\ln \left|\csc \left(\frac{x}{2}+\frac{\pi}{4} \right) + \cot \left(\frac{x}{2}+\frac{\pi}{4} \right) \right| +c.$

Interesting approach, here's my approach:

$\int \frac{\sin{x}}{\sqrt{\sin{x}+1}} \times \frac{\sqrt{1-\sin{x}}}{\sqrt{1-\sin{x}}} dx$

$=\int \frac{\sin{x}\sqrt{1-\sin{x}}}{\cos{x}} dx $ Let u $ = \sqrt{1-\sin{x}}$

$= \int \frac{-2(1-u^2)}{2-u^2} du = 2\int \frac{1}{2-u^2} -1 du$

$= \frac{1}{2}ln\left(\frac{2+\sqrt{1-\sin{x}}}{2-\sqrt{1-\sin{x}}}\right) -2\sqrt{1-\sin{x}} +c$

leehuan · May 30, 2015

Re: MX2 2015 Integration Marathon

$\int _{ 0 }^{ \infty }{ { e }^{ -x }\sin { x } \cos { x } dx }$

Drsoccerball · May 30, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int _{ 0 }^{ \infty }{ { e }^{ -x }\sin { x } \cos { x } dx }$

$\int_0^{\infty}{\frac{1}{2} e^{-x}sin2x}dx$
IBP twice and donzo

porcupinetree · Jun 3, 2015

Re: MX2 2015 Integration Marathon

Not sure if anyone's asked this; it's a fun one which requires the use of complex numbers (at least for my method of working it out):

$\int \sqrt{\tan{x}}} dx$

Drsoccerball · Jun 3, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Not sure if anyone's asked this; it's a fun one which requires the use of complex numbers (at least for my method of working it out):

$\int \sqrt{\tan{x}}} dx$

Lee solved this using the substitution of tan inverse u

leehuan · Jun 4, 2015

Re: MX2 2015 Integration Marathon

$\int { \frac { \tan { x } \sqrt { \sec { x } } +\sec { x } \sqrt { \tan { x } } }{ \cos { x } } dx }$

Natural Water · Jun 4, 2015

Re: MX2 2015 Integration Marathon

Solid question leehuan

Drsoccerball · Jun 4, 2015

Re: MX2 2015 Integration Marathon

$\int{xcotx}$

leehuan · Jun 4, 2015

Re: MX2 2015 Integration Marathon

Natural Water said:
Solid question leehuan

View attachment 32101

Neat first post.

Drsoccerball said:
$\int{xcotx}$

Can't be evaluated by standard methods.
------------
Not sure if Ext 2 maths suffices for this, however trialling with values for 'n' gave an interesting result

$\int { \frac { dx }{ x\left( x+1 \right) \left( x+2 \right) \left( x+3 \right) \dots \left( x+n \right) } } \quad n\in \mathbb Z$

Drsoccerball · Jun 4, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Neat first post.

Can't be evaluated by standard methods.
------------
Not sure if Ext 2 maths suffices for this, however trialling with values for 'n' gave an interesting result

$\int { \frac { dx }{ x\left( x+1 \right) \left( x+2 \right) \left( x+3 \right) \dots \left( x+n \right) } } \quad n\in \mathbb Z$

Is the answer

$\sum_{k=0}^{n}{ln|x+k| \cdot \frac{(-1)^{k}}{(n-k)!}} +C$

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