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

braintic · Sep 27, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Plenty. A classic example is the below, where the limit of the derivative does not exist, but it most certainly has a horizontal asymptote.

$f(x) = \frac{\sin(x^2)}{x}$

That would depend on how you define an asymptote.
I believe some sources actually define it as the limiting tangent.

Carrotsticks · Sep 28, 2015

Re: MX2 2015 Integration Marathon

braintic said:
That would depend on how you define an asymptote.
I believe some sources actually define it as the limiting tangent.

If we define asymptotes to be limiting tangents, then the curve I have mentioned above will have no asymptote.

What would be the significance of the line y=0 then? Clearly it has some significance, but it cannot be called an asymptote.

Paradoxica · Sep 28, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
If we define asymptotes to be limiting tangents, then the curve I have mentioned above will have no asymptote.

What would be the significance of the line y=0 then? Clearly it has some significance, but it cannot be called an asymptote.

Limiting value?

Anyway, next question.

$\int \ln(x+\sqrt{x^2+k})dx$

Carrotsticks · Sep 28, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Limiting value?

Anyway, next question.

$\int \ln(x+\sqrt{x^2+k})dx$

Okay, add an X to my example from earlier. What is the significance of the line y=x now?

integral95 · Sep 28, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Limiting value?

Anyway, next question.

$\int \ln(x+\sqrt{x^2+k})dx$

I'm assuming k is just a constant,

using by parts

$u = ln(x+\sqrt{x^2+k^2}) \ \ \ \ \ v' = 1 \\ \\ \\ u' = \frac{1}{\sqrt{x^2+k^2}} \ \ \ \ \ v = x$

$\Rightarrow xln(x+\sqrt{x^2+k^2}) - \int\frac{x}{\sqrt{x^2+k^2}} \ \ dx$

and yeah just let u = x^2+k^2 in the 2nd integral and done.

Drsoccerball · Sep 28, 2015

Re: MX2 2015 Integration Marathon

integral95 said:
I'm assuming k is just a constant,

using by parts

$u = ln(x+\sqrt{x^2+k^2}) \ \ \ \ \ v' = 1 \\ \\ \\ u' = \frac{1}{\sqrt{x^2+k^2}} \ \ \ \ \ v = x$

$\Rightarrow xln(x+\sqrt{x^2+k^2}) - \int\frac{x}{\sqrt{x^2+k^2}} \ \ dx$

and yeah just let u = x^2+k^2 in the 2nd integral and done.

Or use reverse chain rule.

Paradoxica · Sep 28, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Or use reverse chain rule.

care to demonstrate?

psyc1011 · Sep 28, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
care to demonstrate?

sqrt(x) = int of x' / (2*sqrt(x)) dx

Thats formula

Drsoccerball · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
care to demonstrate?

I mean instead of doing the u substitution just do reverse chain rule.

leehuan · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Believe it or not, some people still can't see the reverse chain rule, but nonetheless still seem to find the substitution and breaking down of the integral into a standard form helpful

Paradoxica · Sep 29, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Believe it or not, some people still can't see the reverse chain rule, but nonetheless still seem to find the substitution and breaking down of the integral into a standard form helpful

Upon closer inspection of my notes, I appear to have used what you declare to be "reverse chain rule" without actually stating that...
Well, you can't always label everything....

$\int_0^{\frac{\pi}{2}}{\frac{x\sin{x} \cos{x}}{\sin^4{x}+\cos^4{x}}}dx$

Ekman · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Upon closer inspection of my notes, I appear to have used what you declare to be "reverse chain rule" without actually stating that...
Well, you can't always label everything....

$\int_{0}^{\frac{\pi}{2}} \frac{x\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx$

ftfy

Edit: Here is the solution:

$I=\int_{0}^{\frac{\pi}{2}} \frac{x\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx = \int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x)\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx$

$\therefore 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx$

$I= \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\tan{x}\sec^{2}{x}}{\tan^4{x}+1}dx $ (Divide top and bottom by $ \cos^{4}{x})$

$\therefore I = \frac{\pi}{8} \left[ \tan^{-1}({\tan^{2}{x}}) \right]_{0}^{\frac{\pi}{2}}$

$I = \frac{\pi}{8} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi^2}{16}$

Paradoxica · Sep 29, 2015

Re: MX2 2015 Integration Marathon

$\int \frac{dx}{\sin^6{x}+\cos^6{x}}$

Ekman · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int \frac{dx}{\sin^6{x}+\cos^6{x}}$

$I= \int \frac{1}{\sin^{6}{x} + \cos^{6}{x}} = \int \frac{\sec^{4}{x} \cdot \sec^{2}{x}}{\tan^{6}+1} dx $ (Divide top and bottom by $ \cos^{6}{x} )$

$$ Let u $ =\tan{x} $ $ \therefore I = \int \frac{(u^2+1)^2}{u^6+1} du = \int \frac{(u^2+1)^2}{(u^2+1)(u^4-u^2 + 1)} du = \int \frac{u^2+1}{u^4 - u^2+1} du$

$I = \int \frac{1+\frac{1}{u^2}}{u^2 - 1 + \frac{1}{u^2}} du $ (Divide top and bottom by $ u^2)$

$$ Let $ a = u - \frac{1}{u} $ $ \therefore I = \int \frac{1}{a^2+1} da = \tan^{-1}{(a)} + c$

$\therefore I = \tan^{-1}{(\tan{x} - \cot{x})} +c$

Drsoccerball · Sep 29, 2015

Re: MX2 2015 Integration Marathon

$I did factorisation:$

$\int{\frac{1}{1-3(sin^2xcos^2x)}}dx$

$\frac{1}{2} \int{\frac{1}{1-\sqrt{3}sinx cosx} + \frac{1}{1+ \sqrt{3}sinxcosx}}$

$Then t substitution$
EDIT: My bad realised it gets too complicated.

Paradoxica · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$I did factorisation:$

$\int{\frac{1}{1-3(sin^2xcos^2x)}}dx$

$\frac{1}{2} \int{\frac{1}{1-\sqrt{3}sinx cosx} + \frac{1}{1+ \sqrt{3}sinxcosx}}$

$Then t substitution$
EDIT: My bad realised it gets too complicated.

For questions of similar form involving congruent even powers of sine and cosine, the best thing to do is divide throughout by the power of the cosine and make a tangent substitution. Or if you're masochistic, divide throughout by sine and make a cotangent substitution.

Ekman said:
$I= \int \frac{1}{\sin^{6}{x} + \cos^{6}{x}} = \int \frac{\sec^{4}{x} \cdot \sec^{2}{x}}{\tan^{6}+1} dx $ (Divide top and bottom by $ \cos^{6}{x} )$

$$ Let u $ =\tan{x} $ $ \therefore I = \int \frac{(u^2+1)^2}{u^6+1} du = \int \frac{(u^2+1)^2}{(u^2+1)(u^4-u^2 + 1)} du = \int \frac{u^2+1}{u^4 - u^2+1} du$

$I = \int \frac{1+\frac{1}{u^2}}{u^2 - 1 + \frac{1}{u^2}} du $ (Divide top and bottom by $ u^2)$

$$ Let $ a = u - \frac{1}{u} $ $ \therefore I = \int \frac{1}{a^2+1} da = \tan^{-1}{(a)} + c$

$\therefore I = \tan^{-1}{(\tan{x} - \cot{x})} +c$

Very good, but your answer can be further simplified. Regardless, onwards and upwards!

a) Differentiate $f(x)e^x$

b) Hence or otherwise, solve
$\int{\frac{2-x^2}{(1-x)\sqrt{1-x^2}}e^x}dx$

Ekman · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
For questions of similar form involving congruent even powers of sine and cosine, the best thing to do is divide throughout by the power of the cosine and make a tangent substitution. Or if you're masochistic, divide throughout by sine and make a cotangent substitution.

Very good, but your answer can be further simplified. Regardless, onwards and upwards!

a) Differentiate $f(x)e^x$

b) Hence or otherwise, solve
$\int{\frac{2-x^2}{(1-x)\sqrt{1-x^2}}e^x}dx$

I know that $\tan{x} - \cot{x} = -2\cot{2x}$ but I thought it looked cooler in its unsimplified form.

I am not entirely sure how your new question is solved though. But according to Wolfram, there is no solution for it.

leehuan · Sep 29, 2015

Re: MX2 2015 Integration Marathon

Can't find anything intuitive to arrange into the form ${ e }^{ x }\left( f\left( x \right) +f^{ \prime }\left( x \right) \right) \\$

braintic · Sep 29, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Can't find anything intuitive to arrange into the form ${ e }^{ x }\left( f\left( x \right) +f^{ \prime }\left( x \right) \right) \\$

What is the significance of this form?

Drsoccerball · Sep 29, 2015

Re: MX2 2015 Integration Marathon

braintic said:
What is the significance of this form?

Is it trying to suggest something to do with IBP

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