HSC 2015 MX2 Integration Marathon (archive) (6 Viewers)

Carrotsticks · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Or you can sub in u=e^(kx) and use partial fractions.

ew

But that'll pretty much be the same as the other solution.

Sy123 · Jun 14, 2015

Re: MX2 2015 Integration Marathon

$\\ \int_0^{\pi /2} \frac{dx}{a^2\sin^2x + b^2\cos^2x}$

leehuan · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \int_0^{\pi /2} \frac{dx}{a^2\sin^2x + b^2\cos^2x}$

$\\ Let\quad u=\tan { x } \Rightarrow du=\sec ^{ 2 }{ x } dx\\ x=0\Rightarrow u=0,x\rightarrow \frac { \pi }{ 2 } \Rightarrow u\rightarrow \infty \\ \int _{ 0 }^{ \pi /2 } \frac { dx }{ a^{ 2 }\sin ^{ 2 } x+b^{ 2 }\cos ^{ 2 } x } \\ =\int _{ 0 }^{ \pi /2 } \frac { \sec ^{ 2 }{ x } dx }{ a^{ 2 }\tan ^{ 2 }{ x } +b^{ 2 } } \\ =\frac { 1 }{ { a }^{ 2 } } \int _{ 0 }^{ \infty }{ \frac { du }{ { u }^{ 2 }+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } }$
$\\ =\frac { 1 }{ { a } } \lim _{ k\rightarrow \infty }{ { \left[ \frac { 1 }{ b } \arctan { \left( \frac { au }{ b } \right) } \right] }_{ 0 }^{ k } } \\ =\frac { 1 }{ { a } } \left( \frac { \pi }{ 2 } -\frac { 1 }{ b } \arctan { 0 } \right) \\ =\frac { \pi }{ 2a }$

Drsoccerball · Jun 14, 2015

Re: MX2 2015 Integration Marathon

$\int_0^{\pi/2}{\frac{sinx}{1+sin^2x}}dx$

leehuan · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Another way which may be a little faster is to divide top and bottom by e^(kx) and it's a log integral.

For that question, I would've done same:
$\int { \frac { dx }{ 1-{ e }^{ kx } } } \\ =\int { \frac { { e }^{ -kx } }{ { e }^{ -kx }-1 } dx } \\ =-\frac { 1 }{ k } \log { \left| { e }^{ -kx }-1 \right| } +C$

However, can someone please explain how the arbitrary constant can differ by an $x$ ?

InteGrand · Jun 14, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
For that question, I would've done same:
$\int { \frac { dx }{ 1-{ e }^{ kx } } } \\ =\int { \frac { { e }^{ -kx } }{ { e }^{ -kx }-1 } dx } \\ =-\frac { 1 }{ k } \log { \left| { e }^{ -kx }-1 \right| } +C$

However, can someone please explain how the arbitrary constant can differ by an $x$ ?

They had $\mathrm{e}^{kx}$ in their logarithm, not $\mathrm{e}^{-kx}$ .

kawaiipotato · Jun 14, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
For that question, I would've done same:
$\int { \frac { dx }{ 1-{ e }^{ kx } } } \\ =\int { \frac { { e }^{ -kx } }{ { e }^{ -kx }-1 } dx } \\ =-\frac { 1 }{ k } \log { \left| { e }^{ -kx }-1 \right| } +C$

However, can someone please explain how the arbitrary constant can differ by an $x$ ?

It doesn't, collect the inside of your log function and use log rules to take out the division. you get something like + 1/k ln(e^kx) which simplifies to be x

leehuan · Jun 14, 2015

Re: MX2 2015 Integration Marathon

24/7 Lost in algebra.

$\\ Let\quad u=\sec { x } \Rightarrow du=\sec { x } \tan { x } \\ x=0\Rightarrow u=1,\quad x\rightarrow \frac { \pi }{ 2 } \Rightarrow u\rightarrow \infty \\ \int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } }{ 1+\sin ^{ 2 }{ x } } } dx\\ =\int _{ 0 }^{ \pi /2 }{ \frac { \sec { x } \tan { x } }{ \sec ^{ 2 }{ x } +\tan ^{ 2 }{ x } } } dx\\ =\int _{ 0 }^{ \pi /2 }{ \frac { \sec { x } \tan { x } }{ 2\sec ^{ 2 }{ x } -1 } } dx\\ =\int _{ 1 }^{ \infty }{ \frac { du }{ 2{ u }^{ 2 }-1 } }$
$I\quad hate\quad partial\quad fractions...\\ =\frac { 1 }{ 2\sqrt { 2 } } \int _{ 1 }^{ \infty }{ \left( \frac { 1 }{ u-\frac { 1 }{ \sqrt { 2 } } } -\frac { 1 }{ u+\frac { 1 }{ \sqrt { 2 } } } \right) } du\\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| u-\frac { 1 }{ \sqrt { 2 } } \right| } -\log { \left| u+\frac { 1 }{ \sqrt { 2 } } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| \frac { u-\frac { 1 }{ \sqrt { 2 } } }{ u+\frac { 1 }{ \sqrt { 2 } } } \right| } \right] }_{ 1 }^{ \infty } }$
$\\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| 1-\frac { \frac { 2 }{ \sqrt { 2 } } }{ u+\frac { 1 }{ \sqrt { 2 } } } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| 1-\frac { 2 }{ \sqrt { 2 } u+1 } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \left( \log { \left| 1-0 \right| } -\log { \left| 1-\frac { 2 }{ \sqrt { 2 } +1 } \right| } \right) \\ =-\frac { 1 }{ 2\sqrt { 2 } } \log { \left( 3-2\sqrt { 2 } \right) }$

Edit #2: This could've sped up the start I suppose.
$\int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } }{ 1+\sin ^{ 2 }{ x } } } dx\\ =-\int _{ 0 }^{ \pi /2 }{ \frac { -\sin { x } }{ 2-\cos ^{ 2 }{ x } } } dx\\ =-\int _{ 1 }^{ 0 }{ \frac { du }{ 2-{ u }^{ 2 } } }$
Edit #3: And everything else, since there's no infinite boundary

porcupinetree · Jun 14, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
24/7 Lost in algebra.

$\\ Let\quad u=\sec { x } \Rightarrow du=\sec { x } \tan { x } \\ x=0\Rightarrow u=1,\quad x\rightarrow \frac { \pi }{ 2 } \Rightarrow u\rightarrow \infty \\ \int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } }{ 1+\sin ^{ 2 }{ x } } } dx\\ =\int _{ 0 }^{ \pi /2 }{ \frac { \sec { x } \tan { x } }{ \sec ^{ 2 }{ x } +\tan ^{ 2 }{ x } } } dx\\ =\int _{ 0 }^{ \pi /2 }{ \frac { \sec { x } \tan { x } }{ 2\sec ^{ 2 }{ x } -1 } } dx\\ =\int _{ 1 }^{ \infty }{ \frac { du }{ 2{ u }^{ 2 }-1 } }$
$I\quad hate\quad partial\quad fractions...\\ =\frac { 1 }{ 2\sqrt { 2 } } \int _{ 1 }^{ \infty }{ \left( \frac { 1 }{ u-\frac { 1 }{ \sqrt { 2 } } } -\frac { 1 }{ u+\frac { 1 }{ \sqrt { 2 } } } \right) } du\\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| u-\frac { 1 }{ \sqrt { 2 } } \right| } -\log { \left| u+\frac { 1 }{ \sqrt { 2 } } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| \frac { u-\frac { 1 }{ \sqrt { 2 } } }{ u+\frac { 1 }{ \sqrt { 2 } } } \right| } \right] }_{ 1 }^{ \infty } }$
$\\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| 1-\frac { \frac { 2 }{ \sqrt { 2 } } }{ u+\frac { 1 }{ \sqrt { 2 } } } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \lim _{ k\rightarrow \infty }{ { \left[ \log { \left| 1-\frac { 2 }{ \sqrt { 2 } u+1 } \right| } \right] }_{ 1 }^{ \infty } } \\ =\frac { 1 }{ 2\sqrt { 2 } } \left( \log { \left| 1-0 \right| } -\log { \left| 1-\frac { 2 }{ \sqrt { 2 } +1 } \right| } \right) \\ =-\frac { 1 }{ 2\sqrt { 2 } } \log { \left( 3-2\sqrt { 2 } \right) }$

Edit #2: This could've sped up the start I suppose.
$\int _{ 0 }^{ \pi /2 }{ \frac { \sin { x } }{ 1+\sin ^{ 2 }{ x } } } dx\\ =-\int _{ 0 }^{ \pi /2 }{ \frac { -\sin { x } }{ 2-\cos ^{ 2 }{ x } } } dx\\ =-\int _{ 1 }^{ 0 }{ \frac { du }{ 2-{ u }^{ 2 } } dx }$
Edit #3: And everything else, since there's no infinite boundary

Ah I see you mention the method that I used. Here was my working: https://imgur.com/LdJUPbh

Drsoccerball · Jun 14, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{tanx+1}{1-tanx}}$ biggest face palm

Ekman · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{tanx+1}{1-tanx}}$ biggest face palm

$\int{\frac{tanx+1}{1-tanx}} = \int \frac{\sin{x} + \cos{x}}{\cos{x} - \sin{x}} dx$

$= -\ln{(\cos{x} - \sin{x})} +c$

Drsoccerball · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int{\frac{tanx+1}{1-tanx}} = \int \frac{\sin{x} + \cos{x}}{\cos{x} - \sin{x}} dx$

$= -\ln{(\cos{x} - \sin{x})} +c$

or $\int{\frac{tanx+1}{1-tanx}} = \int{tan(x+\frac{\pi}{4}})$

leehuan · Jun 15, 2015

Re: MX2 2015 Integration Marathon

$\int { \frac { 1+\sin { x } +\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } }{ 1+\cos { x } +\cos ^{ 2 }{ x } +\cos ^{ 3 }{ x } } dx }$

mreditor16 · Jun 17, 2015

Re: MX2 2015 Integration Marathon

I realise there is still an unresolved integral above. But really wanted to post this integral, which one of my students gave to me - I'm struggling to find a neat solution for it. Anyone able to help out??

( And, before anyone asks, there is no typo

)

EDIT - Oopsss, forgot the dx

mreditor16 · Jun 17, 2015

Re: MX2 2015 Integration Marathon

Drongoski said:
No typo? What about the "dx" part?

sorry sorry, argh forgot that. :/ :/

InteGrand · Jun 17, 2015

Re: MX2 2015 Integration Marathon

mreditor16 said:
I realise there is still an unresolved integral above. But really wanted to post this integral, which one of my students gave to me - I'm struggling to find a neat solution for it. Anyone able to help out?? ( And, before anyone asks, there is no typo )

EDIT - Oopsss, forgot the dx

The numerator is very close to the derivative of the denominator. You could trying fiddling around with that in mind. (E.g. Replace the 3 in the numerator with -3, and compensate by adding a +6 as a second integral; this second integral can be done more neatly with t method than the first.)

Also, t-method works for the original integral of course, just not too "neat" perhaps.

InteGrand · Jun 17, 2015

Re: MX2 2015 Integration Marathon

mreditor16 said:
I realise there is still an unresolved integral above. But really wanted to post this integral, which one of my students gave to me - I'm struggling to find a neat solution for it. Anyone able to help out?? ( And, before anyone asks, there is no typo )

EDIT - Oopsss, forgot the dx

Another trick is to write the numerator as:

$3\sin x + 2\cos x \equiv A\underbrace{(2\sin x + 3\cos x)}_\text{denominator}+B\underbrace{(-3\sin x + 2\cos x)}_\text{derivative of denominator}$ .

After finding the suitable A and B (by solving a pair of simultaneous equations), the integral will become $\int \left( A + B\frac{D^\prime (x)}{D(x)}\right)\text{ d}x$ , where D(x) is the denominator, which we can easily find.

mreditor16 · Jun 17, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
The numerator is very close to the derivative of the denominator. You could trying fiddling around with that in mind. (E.g. Replace the 3 in the numerator with -3, and compensate by adding a +6 as a second integral; this second integral can be done more neatly with t method than the first.)

Also, t-method works for the original integral of course, just not too "neat" perhaps.

I actually did that, but it is veryyy yuckkkkk. :/

mreditor16 · Jun 17, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
The numerator is very close to the derivative of the denominator. You could trying fiddling around with that in mind. (E.g. Replace the 3 in the numerator with -3, and compensate by adding a +6 as a second integral; this second integral can be done more neatly with t method than the first.)

Also, t-method works for the original integral of course, just not too "neat" perhaps.

InteGrand said:
Another trick is to write the numerator as:

$3\sin x + 2\cos x \equiv A\underbrace{(2\sin x + 3\cos x)}_\text{denominator}+B\underbrace{(-3\sin x + 2\cos x)}_\text{derivative of denominator}$ .

After finding the suitable A and B (by solving a pair of simultaneous equations), the integral will become $\int \left( A + B\frac{D^\prime (x)}{D(x)}\right)\text{ d}x$ , where D(x) is the denominator, which we can easily find.

So when I did the bolded thing in the first quoted post yesterday, I actually just ended up with the situation described in the second quoted post. It still gets quite messy when you proceed onwards from the state of affairs raised in your second post.

InteGrand · Jun 17, 2015

Re: MX2 2015 Integration Marathon

mreditor16 said:
So when I did the bolded thing in the first quoted post yesterday, I actually just ended up with the situation described in the second quoted post. It still gets quite messy when you proceed onwards from the state of affairs raised in your second post.

Really? By $D^\prime(x)$ , I meant the derivative of $D(x)$ . So in the second quoted post, basically only one "tedious" step is required, which is to solve a linear system of equations in two unknowns.

Once we do this, we're pretty much done, because the integral of the $\frac{D^\prime (x)}{D(x)}$ part is just $\ln |D(x)|$ .

i.e. $3\sin x + 2\cos x \equiv A(2\sin x + 3\cos x)+B(-3\sin x + 2\cos x)$ , where $A=\frac{12}{13},B=-\frac{5}{13}$ (assuming I didn't make a silly mistake in solving the simultaneous equations, but you get the idea).

Then

$\int \frac{3\sin x + 2\cos x}{2\sin x + 3\cos x}\text{ d}x=\int \frac{A(2\sin x + 3\cos x)+B(-3\sin x + 2\cos x)}{2\sin x + 3\cos x}\text{ d}x$

$=\int \left(A+B\frac{-3\sin x + 2\cos x}{2\sin x + 3\cos x} \right)\text{ d}x$

$=Ax+B\ln |2\sin x + 3\cos x|+c$ .

I'd say that's not too messy.

HSC 2015 MX2 Integration Marathon (archive) (6 Viewers)

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