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

Sy123 · Dec 7, 2014

Here post integrals for helping 2015 students with integration!

Do not post a question until the last one is answered

Do not post in the other thread because that is locked

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$\int_0^{\pi /6} \sin^4 x \cos^3 x \ dx$

turntaker · Dec 7, 2014

Re: MX2 2015 Integration Marathon

Someone try solving this

Sy123 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

turntaker said:
Someone try solving this

You have to solve my integral first, one question at a time

ymcaec · Dec 7, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int_0^{\pi /6} \sin^4 x \cos^3 x \ dx$

$\,\,\,\int_{0}^{\tfrac{\pi }{6}}{{{\sin }^{4}}x\ {{\cos }^{3}}x\,\,dx} \\ =\int_{0}^{\tfrac{\pi }{6}}{{{\sin }^{4}}x\left( 1-{{\sin }^{2}}x \right)\cos x\,\,dx} \\ =\int_{0}^{\tfrac{\pi }{6}}{\left( {{\sin }^{4}}x-{{\sin }^{6}}x \right)\cos x\,\,dx} \\ =\left[ \frac{{{\sin }^{5}}x}{5}-\frac{{{\sin }^{7}}x}{7} \right]_{0}^{\tfrac{\pi }{6}} \\ =\frac{23}{4480} \\$

turntaker · Dec 7, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
You have to solve my integral first, one question at a time

Oh sorry. Didn't read properly

ymcaec · Dec 7, 2014

Re: MX2 2015 Integration Marathon

turntaker said:
Someone try solving this

$\,\,\,\,\int{\frac{1}{1+{{\sin }^{2}}x}dx} \\ \text{=}\int{\frac{1}{{{\cos }^{2}}x+2{{\sin }^{2}}x}dx} \\ =\int{\frac{{{\sec }^{2}}x}{1+2{{\tan }^{2}}x}dx} \\ \\ \text{Let}\,u=\tan x,\,\,\frac{du}{dx}={{\sec }^{2}}x \\ =\frac{1}{2}\int{\frac{du}{\tfrac{1}{2}+{{u}^{2}}}} \\ =\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \sqrt{2}\tan x \right)+c \\$

turntaker · Dec 7, 2014

Re: MX2 2015 Integration Marathon

ymcaec said:
$\,\,\,\,\int{\frac{1}{1+{{\sin }^{2}}x}dx} \\ \text{=}\int{\frac{1}{{{\cos }^{2}}x+2{{\sin }^{2}}x}dx} \\ =\int{\frac{{{\sec }^{2}}x}{1+2{{\tan }^{2}}x}dx} \\ \\ \text{Let}\,u=\tan x,\,\,\frac{du}{dx}={{\sec }^{2}}x \\ =\frac{1}{2}\int{\frac{du}{\tfrac{1}{2}+{{u}^{2}}}} \\ =\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \sqrt{2}\tan x \right)+c \\$

Nice!! Well done

Sy123 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

$\\ \int_0^1 \frac{dx}{1+e^x}$

integral95 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \int_0^1 \frac{dx}{1+e^x}$

I get the feeling there's a faster way, but anyways

$\\ \\ \int_0^1 \frac{dx}{1+e^x} = \int_0^1 \frac{e^x}{e^x(1+e^x)} \ dx \ \ \ \ \ Let \ \ \ u = e^x \\ \\ \Rightarrow \int_1^{e}\frac{du}{(u)(u+1)} = \int_1^{e} \frac{1}{u} - \frac{1}{u+1} \ du \\ \\ \\ = ln(u) - ln(u+1) = ln(e) - ln(e-1) - 0 + ln(2)$

$\int_{-\frac{\pi}{3}}^{-\frac{\pi}{6}} \ tan^{-1}x + tan^{-1}\frac{1}{x} \ \ dx$

SpiralFlex · Dec 7, 2014

Re: MX2 2015 Integration Marathon

integral95 said:
I get the feeling there's a faster way, but anyways

$\\ \\ \int_0^1 \frac{dx}{1+e^x} = \int_0^1 \frac{e^x}{e^x(1+e^x)} \ dx \ \ \ \ \ Let \ \ \ u = e^x \\ \\ \Rightarrow \int_1^{e}\frac{du}{(u)(u+1)} = \int_1^{e} \frac{1}{u} - \frac{1}{u+1} \ du \\ \\ \\ = ln(u) - ln(u+1) = ln(e) - ln(e-1) - 0 + ln(2)$

$\int_{-\frac{\pi}{3}}^{-\frac{\pi}{6}} \ tan^{-1}x + tan^{-1}\frac{1}{x} \ \ dx$

[Hint for slightly faster method: $e^{-x}$ is involved]

Sy123 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

SpiralFlex said:
[Hint for slightly faster method: $e^{-x}$ is involved]

Another method:

$\frac{1}{1+e^x}= 1 - \frac{e^x}{1+e^x}$

integral95 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

Nice!!, I knew I missed something like that

anyways

integral95 said:
$\int_{-\frac{\pi}{3}}^{-\frac{\pi}{6}} \ tan^{-1}x + tan^{-1}\frac{1}{x} \ \ dx$

RealiseNothing · Dec 7, 2014

Re: MX2 2015 Integration Marathon

integral95 said:
$\int_{-\frac{\pi}{3}}^{-\frac{\pi}{6}} \ tan^{-1}x + tan^{-1}\frac{1}{x} \ \ dx$

Just integrating a constant, ceebs computing it.

Sy123 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

$\int_0^{\pi /2} \frac{1}{1 + \tan^4 x} \ $d$x$

integral95 · Dec 7, 2014

Re: MX2 2015 Integration Marathon

i guess so

$$My question$ \ \ = -\frac{\pi}{2}(\frac{\pi}{6}) = -\frac{\pi^2}{12}$

Sy123 · Dec 8, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int_0^{\pi /2} \frac{1}{1 + \tan^4 x} \ $d$x$

$\\ $i) Show that$ \ \int_0^{a} f(x) \ dx = \int_0^a f(a-x) \ dx$

$\\ $ii) Hence find$ \ \int_0^{\pi /2} \frac{\cos^{4}x}{\sin^4x + \cos^4x} \ dx$

InteGrand · Dec 11, 2014

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int_0^{\pi /2} \frac{1}{1 + \tan^4 x} \ $d$x$

$\text{Let } I = \int_0^{\pi /2} \frac{1}{1 + \tan^4 x} \text{ d}x.$

One the one hand,
$I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan^4 \left ( \frac{\pi}{2}-x \right )} \text{ d}x$

$= \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^4 x} \text{ d}x.$

On the other hand, multiplying the numerator and denominator of the integrand of the original integral by cot⁴ x gives

$I = \int_{0}^{\frac{\pi}{2}}\frac{\cot^4 x}{\cot^4 x +1} \text{ d}x.$

Therefore,

$I+I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^4 x} \text{ d}x + \int_{0}^{\frac{\pi}{2}}\frac{\cot^4 x}{\cot^4 x +1} \text{ d}x.$

$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}\frac{1+\cot^4 x}{1+\cot^4 x} \text{ d}x$

$\Rightarrow I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \text{ d}x \Rightarrow I=\frac{\pi}{4}$

emilios · Dec 11, 2014

Re: MX2 2015 Integration Marathon

InteGrand said:
$\text{Let } I = \int_0^{\pi /2} \frac{1}{1 + \tan^4 x} \text{ d}x.$

One the one hand,
$I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan^4 \left ( \frac{\pi}{2}-x \right )} \text{ d}x$

$= \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^4 x} \text{ d}x.$

On the other hand, multiplying the numerator and denominator of the integrand of the original integral by cot⁴ x gives

$I = \int_{0}^{\frac{\pi}{2}}\frac{\cot^4 x}{\cot^4 x +1} \text{ d}x.$

Therefore,

$I+I = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^4 x} \text{ d}x + \int_{0}^{\frac{\pi}{2}}\frac{\cot^4 x}{\cot^4 x +1} \text{ d}x.$

$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}\frac{1+\cot^4 x}{1+\cot^4 x} \text{ d}x$

$\Rightarrow I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \text{ d}x \Rightarrow I=\frac{\pi}{4}$

that is absolute wizardry

FrankXie · Dec 11, 2014

Re: MX2 2015 Integration Marathon

lol a 100% magic done by substitution. anyone can find a way to play this magic:

$\int_1^3\frac{\ln x}{3+x^2}dx$

Sy123 · Dec 12, 2014

Re: MX2 2015 Integration Marathon

FrankXie said:
lol a 100% magic done by substitution. anyone can find a way to play this magic:

$\int_1^3\frac{\ln x}{3+x^2}dx$

Do you mean a substitution that does it all in one go?

The only thing I can think of at the moment is:

$u = \frac{x}{\sqrt{3}} \Rightarrow \ I_1$

$u = \frac{\sqrt{3}}{x} \Rightarrow \ I_2$

$2I = I_1 + I_2 = \int \dots$

Where the dots are a tan inverse integral which is easy to compute

Is that what you were looking for?

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

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