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HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Thread starter bleakarcher
Start date Aug 5, 2012

Status: Not open for further replies.

U

U MAD BRO

Member

#101

Re: MX2 Integration Marathon

$\int (x-1)(x+1)^{11} dx$

MX2 level =)

asianese

Σ

#102

Re: MX2 Integration Marathon

$\int (x-1)(x+1)^{11} dx \\ = \int x(x+1)^{11}-(x+1)^{11} dx \\ = \int [(x+1)-1](x+1)^{11}-(x+1)^{11} dx \\ = \int (x+1)^{12}-2(x+1)^{11} dx \\ = \frac{(x+1)^{13}}{13}-\frac{(x+1)^{12}}{6} +C$

deswa1

Well-Known Member

#103

Re: MX2 Integration Marathon

^ You could have done it slightly faster by changing (x-1) to (x+1-2) in the opening line but still nice. I love these sort of questions.

asianese

Σ

#104

Re: MX2 Integration Marathon

Ah yes....save the hassle. Thanks for the tip

Next:

$\int \sec^3x dx$

Last edited: Aug 10, 2012

nightweaver066

Well-Known Member

#105

Re: MX2 Integration Marathon

asianese said:
Ah yes....save the hassle. Thanks for the tip

Next:

$\int \sec^3x dx$

Was in my 4U test. looks very simple at first but its not lol

$\int secx d(tanx)$

$= secxtanx - \int secx tan^2x dx$

$= secxtanx - \int (sec^2x - 1) secx dx$

$= secxtanx - \int sec^3x - secx dx$

$\therefore 2\int sec^3x dx = secxtanx + ln|secx + tanx|$

$\int sec^3x dx = \frac{1}{2}(secxtanx + ln|secx + tanx|) + c$

Last edited: Aug 10, 2012

Carrotsticks

Retired

#106

Re: MX2 Integration Marathon

nightweaver066 said:
Was in my 4U test. looks easy but an annoying integral.

It's a very commonly asked problem, since students often think "There's only one function, so I can't use LIATE!" and end up freaking out.

U

U MAD BRO

Member

#107

Re: MX2 Integration Marathon

asianese said:
$\int (x-1)(x+1)^{11} dx \\ = \int x(x+1)^{11}-(x+1)^{11} dx \\ = \int [(x+1)-1](x+1)^{11}-(x+1)^{11} dx \\ = \int (x+1)^{12}-2(x+1)^{11} dx \\ = \frac{(x+1)^{13}}{13}-\frac{(x+1)^{12}}{6} +C$

Neat solution!
I think u can also do it by letting u = x+1 =)

nightweaver066

Well-Known Member

#108

Re: MX2 Integration Marathon

Carrotsticks said:
It's a very commonly asked problem, since students often think "There's only one function, so I can't use LIATE!" and end up freaking out.

lol i more or less think of trig substitutions when a question like that pops up

Carrotsticks

Retired

#109

Re: MX2 Integration Marathon

U MAD BRO said:
Neat solution!
I think u can also do it by letting u = x+1 =)

That is the same as Deswa1's method essentially.

U

U MAD BRO

Member

#110

Re: MX2 Integration Marathon

U MAD BRO said:
$\int_{0}^{1}loglog\left ( \frac{1}{x} \right )dx$

good luck!

Is this the Euler Mascheroni constant?

nightweaver066

Well-Known Member

#111

Re: MX2 Integration Marathon

$\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx, \; $using x = sinu$$

U

U MAD BRO

Member

#112

Re: MX2 Integration Marathon

nightweaver066 said:
$\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx, \; $using x = sinu$$

Can I do this:

$\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)$

and avoid using the substitution?

Last edited: Aug 10, 2012

deswa1

Well-Known Member

#113

Re: MX2 Integration Marathon

U MAD BRO said:
Can I do this:

$\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)$

and avoid using the substitution?

Where are you going to go from here?

U

U MAD BRO

Member

#114

Re: MX2 Integration Marathon

deswa1 said:
Where are you going to go from here?

I'm thinking =(

I have to do tutor homework now, will get back to it once I'm done.

U

U MAD BRO

Member

#115

Re: MX2 Integration Marathon

U MAD BRO said:
Can I do this:

$\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)$

and avoid using the substitution?

If we distribute:
$-\frac{1}{x^3}\left(2 + \frac{1}{x^2 - 1}\right) = -\frac{2}{x^3} - \frac{1}{x^3(x^2 - 1)}$

Can we use this Identity to integrate?

$\int \left( f(x) + g(x) \right) dx = \int f(x) dx + \int g(x) dx$

nightweaver066

Well-Known Member

#116

Re: MX2 Integration Marathon

U MAD BRO said:
If we distribute:
$-\frac{1}{x^3}\left(2 + \frac{1}{x^2 - 1}\right) = -\frac{2}{x^3} - \frac{1}{x^3(x^2 - 1)}$

Can we use this Identity to integrate?

$\int \left( f(x) + g(x) \right) dx = \int f(x) dx + \int g(x) dx$

sure, you can. then partial fractions from there

although follow the question and use the substitution lol

U

U MAD BRO

Member

#117

Re: MX2 Integration Marathon

nightweaver066 said:
sure, you can. then partial fractions from there

although follow the question and use the substitution lol

If it says use a given substitution in MX2 do we have to use it? Do we get 0 if we don't?

nightweaver066

Well-Known Member

#118

Re: MX2 Integration Marathon

U MAD BRO said:
If it says use a substitution in MX2 do we have to use it? Do we get 0 if we don't?

yes if it says "Use.."

you don't have to if it says "Use.. or otherwise.."

barbernator

Active Member

#119

Re: MX2 Integration Marathon

ok I will join. Can someone repost the ones that havnt been done please so i can see whats going on

nightweaver066

Well-Known Member

#120

Re: MX2 Integration Marathon

barbernator said:
ok I will join. Can someone repost the ones that havnt been done please so i can see whats going on

the most recent one i posted

spirals one on the first page i think

all the ones "U MAD BRO" posted up this page

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