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

simpleetal · Apr 16, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ $i) Explain why$ \ \int_0^a f(x) \ dx = \int_0^a f(a-x) \ dx \\ \\ $ii) Hence find$ \ \int_0^{\pi/4} \ln (1 + \tan x) \ dx$

$\ \int_0^a f(x) \ dx = - \int_a^0 f(a-u) \ du = \int_0^a f(a-u) \ du $ $which the same as what's desired by change of variables (use substituation x=a-u) $ \\ $ \ \int_0^{\pi/4} \ln (1 + \tan x) \ dx = \int_0^{\pi/4} ln2 \ dx - \ \int_0^{\pi/4} \ln (1+tanx) \ dx $ \\ $by part i$ \\ $Thus $ I= \int_0^{\pi/4} ln2 \ dx -I $ \\ $ I = \frac{\pi*ln2}{8}$

braintic · Apr 16, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
$\ \int_0^{\pi/4} \ln (1 + \tan x) \ dx = \int_0^{\pi/4} ln2 \ dx - \ \int_0^{\pi/4} \ln (1+tanx) \ dx $ \\ $by part i$

You have left out all the important working required to get that result ... ie. using the tan(A-B) result

leehuan · Apr 17, 2015

Re: MX2 2015 Integration Marathon

$I\quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( 1+\tan { x } \right) } dx } ---(1)\\ \therefore I\quad =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( 1+\tan { \left( \frac { \pi }{ 4 } -x \right) } \right) } dx } \\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( 1+\frac { 1-\tan { x } }{ 1+1\tan { x } } \right) } dx } \quad as\quad \tan { \frac { \pi }{ 4 } } =1$

$=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( \frac { 2 }{ 1+\tan { x } } \right) } dx } \\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( 2 \right) } dx } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { \left( 1+\tan { x } \right) } dx } \quad as\quad \log { \frac { A }{ B } } =\log { A } -\log { B }$

$I\quad =\quad { \left[ x\ln { 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }-I\\ 2I\quad =\quad \frac { \pi }{ 4 } \ln { 2 } -0\ln { 2 } \\ I\quad =\quad \frac { \pi \ln { 2 } }{ 8 }$

No plus sign at the end

Drsoccerball · Apr 17, 2015

Re: MX2 2015 Integration Marathon

$\int{\sqrt{tanx}}$

leehuan · Apr 17, 2015

Re: MX2 2015 Integration Marathon

$x=\tan ^{ -1 }{ u } \quad \Rightarrow \quad dx=\frac { du }{ 1+{ u }^{ 2 } }$
$\int { \sqrt { \tan { x } } dx }$
$=\int { \frac { \sqrt { u } }{ 1+{ u }^{ 2 } } du }$
$=\int { \frac { 2u }{ 2\sqrt { u } \left( 1+{ u }^{ 2 } \right) } du }$
$s=\sqrt { u } \quad \Rightarrow \quad ds=\frac { du }{ 2\sqrt { u } }$
$=\int { \frac { 2{ s }^{ 2 } }{ { s }^{ 4 }+1 } ds }$
$Partial\quad fractions\quad with\quad \frac { 2{ s }^{ 2 } }{ { s }^{ 4 }+1 } \equiv \frac { As+B }{ { s }^{ 2 }-\sqrt { 2 } s+1 } +\frac { Cs+D }{ { s }^{ 2 }+\sqrt { 2 } s+1 } \quad turns\quad the\quad integral\quad to$
$=\int { \left( \frac { s }{ \sqrt { 2 } \left( { s }^{ 2 }-\sqrt { 2 } s+1 \right) } +\frac { -s }{ \sqrt { 2 } \left( { s }^{ 2 }+\sqrt { 2 } s+1 \right) } \right) } ds$
$=\frac { 1 }{ 2\sqrt { 2 } } \int { \left( \frac { 2s }{ { s }^{ 2 }-\sqrt { 2 } s+1 } -\frac { 2s }{ { s }^{ 2 }+\sqrt { 2 } s+1 } \right) } ds$
$=\frac { 1 }{ 2\sqrt { 2 } } \int { \left[ \frac { 2s-\sqrt { 2 } }{ { s }^{ 2 }-\sqrt { 2 } s+1 } +\frac { \sqrt { 2 } }{ { \left( s+\frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+\frac { 1 }{ 2 } } -\frac { 2s+\sqrt { 2 } }{ { s }^{ 2 }+\sqrt { 2 } s+1 } +\frac { \sqrt { 2 } }{ { \left( s-\frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+\frac { 1 }{ 2 } } \right] } ds$
$=\frac { 1 }{ 2\sqrt { 2 } } \left[ \ln { \left( { s }^{ 2 }-\sqrt { 2 } s+1 \right) } +\sqrt { 2 } \left( \sqrt { 2 } \tan ^{ -1 }{ \left( \sqrt { 2 } s+1 \right) } \right) -\ln { \left( { s }^{ 2 }+\sqrt { 2 } s+1 \right) } +\sqrt { 2 } \left( \sqrt { 2 } \tan ^{ -1 }{ \left( \sqrt { 2 } s-1 \right) } \right) \right] +C$
$=\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( { s }^{ 2 }-\sqrt { 2 } s+1 \right) } +\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2 } s+1 \right) } -\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( { s }^{ 2 }+\sqrt { 2 } s+1 \right) } +\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2 } s-1 \right) } +C$
$=\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( u-\sqrt { 2u } +1 \right) } +\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2u } +1 \right) } -\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( u+\sqrt { 2u } +1 \right) } +\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2u } -1 \right) } +C$
$=\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( \tan { x } -\sqrt { 2\tan { x } } +1 \right) } \frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2\tan { x } } +1 \right) } -\frac { 1 }{ 2\sqrt { 2 } } \ln { \left( \tan { x } +\sqrt { 2\tan { x } } +1 \right) } +\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \sqrt { 2\tan { x } } -1 \right) } +C$
$=\frac { 1 }{ 2\sqrt { 2 } } \left( \ln { \left( \frac { \tan { x } -\sqrt { 2\tan { x } } +1 }{ \tan { x } +\sqrt { 2\tan { x } } +1 } \right) } +2\tan ^{ -1 }{ \left( \sqrt { 2\tan { x } } +1 \right) } +2\tan ^{ -1 }{ \left( \sqrt { 2\tan { x } } -1 \right) } \right) +C$

VBN2470 · Apr 17, 2015

Re: MX2 2015 Integration Marathon

^ I think using the direct substitution $u^2 = \tan(x)$ would have made the first part a bit easier (no need for a second substitution). Rest looks fine though

swagswagyoloswag · Apr 17, 2015

Re: MX2 2015 Integration Marathon

Jk

leehuan · Apr 17, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
^ I think using the direct substitution $u^2 = \tan(x)$ would have made the first part a bit easier (no need for a second substitution). Rest looks fine though

I would like to agree

${ u }^{ 2 }=\tan { x } \Rightarrow 2udu=\sec ^{ 2 }{ x } dx=\left( 1+\tan ^{ 2 }{ x } \right) dx\\ dx=\frac { 2udu }{ 1+\tan ^{ 2 }{ x } } =\frac { 2udu }{ 1+{ u }^{ 4 } } \\ \int { \sqrt { \tan { x } } dx } \\ =\int { u\frac { 2udu }{ 1+{ u }^{ 4 } } } \\ =\int { \frac { 2{ u }^{ 2 } }{ 1+{ u }^{ 4 } } du } \\$

VBN2470 · Apr 17, 2015

Re: MX2 2015 Integration Marathon

$$ For $n\,>\,1$, evaluate $\int_{0}^{\infty} \frac{\text{ d}x}{(x+\sqrt{1+x^2})^n}$.$

FrankXie · Apr 17, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
$$ For $n\,>\,1$, evaluate $\int_{0}^{\infty} \frac{\text{ d}x}{(x+\sqrt{1+x^2})^n}$.$

substitute $x=\tan\theta$ , then carry out IBP twice. i got $I=\frac{n}{n^2-1}$

swagswagyoloswag · Apr 17, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
substitute $x=\tan\theta$ , then carry out IBP twice

Won't the first IBP be u = 1/(tantheta + sectheta)^n and dv = sec^2 theta

FrankXie · Apr 17, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
Won't the first IBP be u = 1/(tantheta + sectheta)^n and dv = sec^2 theta

yes thats same as i did

swagswagyoloswag · Apr 17, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
yes thats same as i did

but the borders will be from pi/4 to pi/2 and when you get uv, you have to sub theta = pi/2 and that gives infinity

FrankXie · Apr 18, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
but the borders will be from pi/4 to pi/2 and when you get uv, you have to sub theta = pi/2 and that gives infinity

no, at pi/2, you still get finite, actually zero ( by multiplying cos^n both numerator and denominator )

VBN2470 · Apr 18, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
$$ For $n\,>\,1$, evaluate $\int_{0}^{\infty} \frac{\text{ d}x}{(x+\sqrt{1+x^2})^n}$.$

Using the substitution $$u = \ln({x+\sqrt{x^2+1}})$ \,\Rightarrow\, $x$ = $\frac{e^{u}-e^{-u}}{2}$$ should work too, you will still need to use IBP twice to get the desired value.

Drsoccerball · Apr 18, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Using the substitution $$u = \ln({x+\sqrt{x^2+1}})$ \,\Rightarrow\, $x$ = $\frac{e^{u}-e^{-u}}{2}$$ should work too, you will still need to use IBP twice to get the desired value.

cosh and sinh?

VBN2470 · Apr 18, 2015

Re: MX2 2015 Integration Marathon

^ Yep, $$u = \sinh^{-1}x$.$

Drsoccerball · Apr 18, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{lnx}{(1+lnx)^2}}dx$

VBN2470 · Apr 18, 2015

Re: MX2 2015 Integration Marathon

Use the "plus 1 minus 1" trick to split the integrand and apply IBP once to get $$\frac{x}{1+\ln(x)}$ + $C$$

Ekman · Apr 18, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{1}^{2} \frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}-\sqrt{ln(x+3)}}dx$

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