HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
This one seems dodgy lol. For one thing, the cos 2x under the square root in the denominator goes negative at some points in the domain...

Isnt it always positive from 0 to 1?

braintic · Apr 5, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
Isnt it always positive from 0 to 1?

It is zero for x = pi/4, which is 0.78.
It is negative after that.

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

The indefinite integral I was able to get to was:

$\sin^{-1}(u) + c$

Where u = tanx

If I change the boundaries such that it fits with the square root sign (i.e. between 0 to 0.5), I can get a proper definite solution.

Drsoccerball · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
The indefinite integral I was able to get to was:

$\sin^{-1}(u) + c$

Where u = tanx

If I change the boundaries such that it fits with the square root sign (i.e. between 0 to 0.5), I can get a proper definite solution.

Tells me not to go on the X2 integration forum anymore...is the most active member :L

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Tells me not to go on the X2 integration forum anymore...is the most active member :L

...what?

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find $\int_{0}^{\frac{\pi}{4}} \frac{\sec{x}}{\sqrt{\cos2x}} \text{ d}x$$

EDIT. Apologies to all who attempted this, the upper limit is pi/4 not 1, too much posting of integration questions today has f*cked my brain up..

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
EDIT. Apologies to all who attempted this, the upper limit is pi/4 not 1, too much posting of integration questions today has f*cked my brain up..

Then the answer should be $\frac{\pi}{2}$

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Then the answer should be $\frac{\pi}{2}$

Your method?

EDIT: Never mind, I saw that you substituted t = tan(x), which is a neat trick for this q

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Your method?

$\int_{0}^{\frac{\pi}{4}} \frac{\sec(x)}{\sqrt{\cos(2x)}} dx = \int_{0}^{\frac{\pi}{4}}\frac{\sec^2(x)}{\sqrt{1-\tan^2(x)}} dx$ (Just did t-method for cos(2x))

$$ Let u$=\tan(x) \therefore \int_{0}^{1} \frac{1}{\sqrt{1-u^2}} du = [\sin^{-1}(u)]_{0}^{1}=\frac{\pi}{2}$

Drsoccerball · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int_{0}^{\frac{\pi}{4}} \frac{\sec(x)}{\sqrt{\cos(2x)}} dx = \int_{0}^{\frac{\pi}{4}}\frac{\sec^2(x)}{\sqrt{1-\tan^2(x)}} dx$ (Just did t-method for cos(2x))

$$ Let u$=\tan(x) \therefore \int_{0}^{1} \frac{1}{\sqrt{1-u^2}} du = [\sin^{-1}(u)]_{0}^{1}=\frac{\pi}{2}$

T method

? easier way:

$\int \frac{sec(x)}{\sqrt{cos^2x -sin^2x}}$

$Take out $ cos^2 $ From the bottom$

$= \int\frac{sec^2x}{\sqrt{1-tan^2(x)}}$

Drsoccerball · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
T method ? easier way:

$\int \frac{sec(x)}{\sqrt{cos^2x -sin^2x}}$

$Take out $ cos^2 $ From the bottom$

$= \int\frac{sec^2x}{\sqrt{1-tan^2(x)}}$

Oh woops your ones easier

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \sqrt{1+e^x} dx$

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \sqrt{1+e^x} dx$

Asked and answered earlier in this thread:

InteGrand said:
New question

$\text{Find } I\equiv \int \sqrt{1+\text{e}^x}\text{ d}x.$

HeroicPandas said:
Let $u^2 = 1+e^x$ and a little side-note that $1+e^x > 0$ so $|u| > 0$

So, $2udu = e^x dx$

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

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

$I = 2 \int ( 1 - \frac{0.5}{u+1} + \frac{0.5}{u-1} )du$

$I = 2 \sqrt{1+e^x} - ln(1+\sqrt{1+e^x}) + ln( \sqrt{1+e^x} +1) +C$

Question:

$$Find $I = \int \frac{sec^2 (x)}{9 + tan^2 (x)} dx$

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Asked and answered earlier in this thread:

Whoops, my bad. I didnt see it before on the thread

Sy123 · Apr 6, 2015

Re: MX2 2015 Integration Marathon

$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \ dx$

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \ dx$

$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \ dx=\int \frac{\tan x \sec^2x}{1+tan^4x} dx$

$$ Let u$=\tan x \therefore \int \frac{u}{1+u^4} du = \int \frac{u}{1+(u^2)^2} du$

$=\frac{1}{2} \tan^{-1}(u^2) + c =\frac{1}{2} \tan^{-1}(\tan^2x)+c$

Sy123 · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \ dx=\int \frac{\tan x \sec^2x}{1+tan^4x} dx$

$$ Let u$=\tan x \therefore \int \frac{u}{1+u^4} du = \int \frac{u}{1+(u^2)^2} du$

$=\frac{1}{2} \tan^{-1}(u^2) + c =\frac{1}{2} \tan^{-1}(\tan^2x)+c$

Nice solution, I was thinking of doing something like:

$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \ dx = \frac{1}{2}\int \frac{\sin (2x)}{1 - \frac{1}{2}\sin^2(2x)} \ dx = \dots$

But it's a little bit longer I think

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{2x-3}{\sqrt{2-x^4+6x^3-9x^2}}}dx$ would never have saw how to do this in the exam

Sy123 · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{2x-3}{\sqrt{2-x^4+6x^3-9x^2}}}dx$ would never have saw how to do this in the exam

$\\ = \int \frac{2x-3}{\sqrt{2 - (x^2-3x)^2}} \ dx \\ \\ $So, use$ \ u = x^2 -3x$

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{1}{\sqrt{x}-1}}dx$

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