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

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x$$

$\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2}dx = \int_{0}^{\infty} \frac{2x+1}{x^4+2x^3+3x^2+2x+2} dx$

$$ Hence: $ \int_{0}^{\infty} \frac{1}{x^2+1} - \frac{1}{x^2+2x+2} dx$

$= [\tan^{-1}(x) - \tan^{-1}(x+1)]_{0}^{\infty}= \frac{\pi}{4}$

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2}dx = \int_{0}^{\infty} \frac{2x+1}{x^4+2x^3+3x^2+2x+2} dx$

$$ Hence: $ \int_{0}^{\infty} \frac{1}{x^2+1} - \frac{1}{x^2+2x+2} dx$

$= [\tan^{-1}(x) - \tan^{-1}(x+1)]_{0}^{\infty}= \frac{\pi}{4}$

How'd you get the first integral to be equal to the 2nd one?

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
How'd you get the first integral to be equal to the 2nd one?

3U factorisation, and the cancellation of factors of both numerator and denominator
For the second line, I used partial fractions, and the rest is pretty straightforward.

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x$$

ALTERNATE METHOD:

$$ \begin{align*} \int_{0}^{\infty} \frac{2x^3-3x^2+1}{x^6-2x^3+x^2-2x+2} \text{ d}x &= \int_{0}^{\infty} \frac{(2x+1)(x-1)^2}{(x^3-1)^2+(x-1)^2} \text{ d}x \\ &= \int_{0}^{\infty} \frac{(2x+1)(x-1)^2}{(x-1)^2((x^2+x+1)^2+1)} \text{ d}x \\ &= \int_{0}^{\infty} \frac{(2x+1)}{(x^2+x+1)^2+1} \text{ d}x \\ &= [\tan^{-1}(x^2+x+1)]_{0}^{\infty} \\ &= \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4} \end{align*}$

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Find $\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x$$

FrankXie · Apr 5, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$\int \frac{1+x}{\sqrt{1-x^{2}}} dx \\ \frac{-1}{2} \int {\frac{-2}{\sqrt{1-x^{2}} + \frac{-2x}{\sqrt{1-x^{2}}} \\ {\frac{-1}{2}} (-2sin^{-1}(x) + 2 \sqrt{1-x^{2}})$

swagswagyoloswag said:
pls someone fix my eqn. im gonna cry with latex why is it invalid

just to fix the latex problem

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

is this what you wanted ?

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x$$

$$ Let u = -x $ \\ $ du = -dx $ \\ I = \int_{-1}^{1}{tan^{-1}(e^{-u})} du \\ \int_{-1}^{1} tan^{-1}(e^{-x}) dx \\ 2I = \int_{-1}^{1} tan^{-1}(e^x) + tan^{-1}(e^{-x}) dx \\ 2I = \int_{-1}^{1} \frac{\pi}{2} dx \\ 2I = \pi \\ I = \frac{\pi}{2}$

Carrotsticks · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x$$

Let u=-x then use the old 2I= .... trick as well as atan(f)+atan(1/f)=pi/2.

Looks like it becomes pi/2.

EDIT: Never mind, beaten above ^

RealiseNothing · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x$$

Let $u=-x$ and we get:

$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-x}) \text{ d}x$

But we also know that:

$\tan^{-1}(e^x) + \tan^{-1}(e^{-x}) = \frac{\pi}{2}$

So our answer is $\frac{\pi}{2}$

EDIT: Never mind, beaten above twice ^^

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

RealiseNothing said:
Let $u=-x$ and we get:

$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-x}) \text{ d}x$

Can someone please explain this

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Can someone please explain this

He just used the substitution u = -x to obtain the result. I think he skipped some of the steps due to cbf typing on LaTeX.

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION

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

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
He just used the substitution u = -x to obtain the result. I think he skipped some of the steps due to cbf typing on LaTeX.

I understood the substitution to get this:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-u})du$

But what I don't get is how you could draw a conclusion to:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-x})dx$

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
I understood the substitution to get this:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-u})du$

But what I don't get is how you could draw a conclusion to:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-x})dx$

You can also think of it graphically: replacing $x$ with $-x$ has the effect of flipping the graph about the y-axis, so keeping the bounds at -1 to +1 while flipping the graph means the area under the curve is the same.

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
You can also think of it graphically: replacing $x$ with $-x$ has the effect of flipping the graph about the y-axis, so keeping the bounds at -1 to +1 while flipping the graph means the area under the curve is the same.

Any way of showing it algebraically?

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
I understood the substitution to get this:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-u})du$

But what I don't get is how you could draw a conclusion to:
$\int_{-1}^{1} \tan^{-1}(e^x) \text{ d}x = \int_{-1}^{1} \tan^{-1}(e^{-x})dx$

Oh, I see. Basically for any definite integral, it doesn't matter what our variable is, whether it is t or x or even u, since we have just assigned a name to it, so $\int_{1}^{e} \ln{x} \text{ d}x = \int_{1}^{e} \ln{t} \text{ d}t = 1$ (see how it doesn't matter if it is t or x, the value remains the same, think of it as being like a 'dummy' variable.

Ekman · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Oh, I see. Basically for any definite integral, it doesn't matter what our variable is, whether it is t or x or even u, since we have just assigned a name to it, so $\int_{1}^{e} \ln{x} \text{ d}x = \int_{1}^{e} \ln{t} \text{ d}t = 1$ (see how it doesn't matter if it is t or x, the value remains the same, think of it as being like a 'dummy' variable.

So then this would only work for cases when there are definite integrals?

VBN2470 · Apr 5, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
So then this would only work for cases when there are definite integrals?

For the indefinite case, if there is a substitution, it won't really work then you would need to be careful with the variables presented. The definite integral is always independent of the variable of integration, so it doesn't matter even if you use a substitution, you will end up getting the same value. Again, it just comes down to the naming of the variables, the mathematical argument is still retained.

swagswagyoloswag · Apr 5, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
just to fix the latex problem

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

is this what you wanted ?

yep lol. thanks

InteGrand · Apr 5, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find $\int_{0}^{1} \frac{\sec{x}}{\sqrt{\cos2x}} \text{ d}x$$

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...

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