MX2 Integration Marathon (2 Viewers)

fan96 · Sep 8, 2018

Re: HSC 2018 MX2 Integration Marathon

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

Are you sure the integrand is correct?

HeroWise · Sep 8, 2018

Re: HSC 2018 MX2 Integration Marathon

Lets campaign the teachers to give mrbunton the mark!

fan96 · Sep 9, 2018

Re: HSC 2018 MX2 Integration Marathon

In the case of a misprinted question, I think the best thing to do here would just be to give everyone the marks.

jyu · Sep 9, 2018

Re: HSC 2018 MX2 Integration Marathon

si2136 said:
Surely it's been written wrong? Only way I can think of using Taylor, but that is outside 4U scope.

Try replacing both e^x with e^(x^2).

Paradoxica · Sep 12, 2018

Re: HSC 2018 MX2 Integration Marathon

The above question is literally impossible, in the sense there is no elementary closed form in terms of a finite combination of elementary operations and HSC-level constants.

juantheron · Oct 23, 2018

Re: HSC 2018 MX2 Integration Marathon

$Let $I = \int^{1}_{0}\frac{\ln(1+x)}{1+x^2}dx,$

$$Put $x=\frac{1-t}{1+t}=\frac{2}{t+1}-1\;,$ Then $dx=-\frac{2}{(1+t)^2}dt$

$And changing limits$

$So $I = \int^{1}_{0}\frac{\ln(2)-\ln(1+t)}{1+t^2}dt=\int^{1}_{0}\frac{\ln(2)}{1+t^2}dt-I$

$So we have $2I = \frac{\pi}{4}\cdot \ln(2)\Rightarrow I = \frac{\pi}{8}\cdot \ln(2).$

stupid_girl · Oct 28, 2018

Re: HSC 2018 MX2 Integration Marathon

Show that
$\int_{0}^{\frac{\pi}{4}}\frac{2\sin 8x \cos 4x \cos 2x}{(1+\cos 8x)(1+\cos 4x)(1+\cos 2x))\sqrt{2+\cos 2x}}dx$

$=\ln(6+4\sqrt{2}-3\sqrt{3}-2\sqrt{6})$

fan96 · Nov 11, 2018

Re: HSC 2018 MX2 Integration Marathon

$\int^{\pi/4}_0 \frac{2 \sin 8x \cos 4x \cos 2x}{(1 + \cos 8x)(1+ \cos 4x)(1 + \cos 2x) \sqrt{2 + \cos 2x}}\, dx$

Using the identities:

$1 + \cos kx = 2 \cos ^2 (kx/2)$

$\sin kx = 2 \sin (kx/2) \cos (kx/2)$

the integral reduces to:

$\int^{\pi/4}_0 \frac{2\tan x}{\sqrt{2 + \cos 2x}}\, dx$

$= 2 \int^{\pi/4}_0 \frac{1}{\cos x\sqrt{1 + 2 \cos^2x}} \cdot \sin x\, dx$

Perform a substitution $u = \cos x$ to get

$= -2 \int^{\sqrt 2/2}_1 \frac{1}{u \sqrt{1 + 2u^2}} \, du$

The substitution $v = \sqrt{1 + 2u^2}$ would probably have given a quicker answer, but the first thing that came to mind was another trig sub:

$u = \frac{1}{\sqrt2} \tan \theta$

Giving:

$-2 \int^{\pi/4}_{\arctan \sqrt 2} \frac{\frac{1}{\sqrt 2} \sec^2 \theta}{\frac{1}{\sqrt2} \tan \theta | \sec \theta | }\, d\theta$

Noting that $- \pi/2 <\arctan \alpha < \pi/2$ for all real $\alpha$ and so $| \sec \theta | = \sec \theta$ for the bounds of this integral, we may simplify to get

$-2 \int_{\arctan \sqrt 2}^{\pi/4} \csc \theta \, d\theta$

$= -2 \Big[-\log | \csc \theta + \cot \theta| \Big]_{\arctan \sqrt 2}^{\pi/4}$

Note: $\csc \arctan \sqrt 2$ and $\cot \arctan \sqrt 2$ can be evaluated by using an appropriate right angled triangle.

$= 2 \log \left| \sqrt 2 + 1 \right|-2 \log \left| \frac{\sqrt 3}{\sqrt 2} + \frac{1}{\sqrt 2} \right|$

$= \log\left[ \left(\frac{2 + \sqrt 2}{\sqrt 3 + 1} \right)^2\right]$

Expanding the square and rationalising the denominator gives

$\log \left(6 - 3 \sqrt 3+ 4 \sqrt 2 - 2 \sqrt 6\right)$

Paradoxica · Nov 11, 2018

Re: HSC 2018 MX2 Integration Marathon

Continuing on from the simplification

$\int_0^\frac{\pi}{4} \frac{2\tan{x}}{\sqrt{2+\cos{2x}}}\text{d}x = \int_0^\frac{\pi}{4} \frac{2\sin{2x}}{(1+\cos{2x} )\sqrt{2+\cos{2x}}}\text{d}x$

Reverse the chain rule twice to obtain:

$\int_0^\frac{\pi}{4} \frac{-2\text{d}\left(\sqrt{2+\cos{2x}}\right)}{(1+\cos{2x})}$

Complete the substitution to obtain:

$\int_{\sqrt{2}}^{\sqrt{3}} \frac{2 \text{d}z}{z^2-1}$

Using the former table of standard integrals, the integral evaluates to:

$\log{\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)} - \log{\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)}$

Which "simplifies" to:

$2\log{\left(\sqrt{2}+1\right)}+2\log{\left(\sqrt{3}-1\right)}-\log{2}$

fan96 · Nov 21, 2018

Re: HSC 2018 MX2 Integration Marathon

Maybe something a bit easier:

$\int_0^\pi \frac{1}{3 + 2\cos x}\,dx$

HeroWise · Nov 21, 2018

Re: HSC 2018 MX2 Integration Marathon

Damn that pi lemme sit on it a n=bit longer

stupid_girl · Dec 9, 2018

Re: HSC 2018 MX2 Integration Marathon

Show that
$\int_{0}^{\frac{\pi}{2}} \frac{sin(x+\frac{\pi}{4})}{(2^\pi+16^x)(sin^3x+cos^3x)}dx=\frac{\sqrt{6}\pi}{9(2^\pi)}$

stupid_girl · Dec 9, 2018

Re: HSC 2018 MX2 Integration Marathon

HeroWise said:
Damn that pi lemme sit on it a n=bit longer

Just consider the integral from 0 to a first, then take the limit as a->pi^-.

fan96 · Dec 9, 2018

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
Show that
$\int_{0}^{\frac{\pi}{2}} \frac{sin(x+\frac{\pi}{4})}{(2^\pi+16^x)(sin^3x+cos^3x)}dx=\frac{\sqrt{6}\pi}{9(2^\pi)}$

If you don't mind me asking, where do you get these integrals from?

fan96 · Dec 10, 2018

Re: HSC 2018 MX2 Integration Marathon

I put my solution as an attachment so that it won't spoil the answer for other people attempting to solve this.

stupid_girl · Dec 10, 2018

Re: HSC 2018 MX2 Integration Marathon

fan96 said:
If you don't mind me asking, where do you get these integrals from?

Construct from simpler integrals using trig identities and properties of definite integrals.

stupid_girl · Dec 15, 2018

Re: HSC 2018 MX2 Integration Marathon

Continue to have fun with trig.

Harder version:
$f(x)=\frac{\sin(\frac{\pi}{2}\sqrt{x})}{\sqrt{2}+ \sqrt{2}\cos(\frac{\pi}{2}\sqrt{x})}+\frac{x\sec (\frac{\pi}{4})}{ \tan(\frac{\pi}{4}\sqrt{1-x})+1}-\frac{x\sin(\frac{\pi}{4}\sqrt{x})}{\cos(\frac{\pi}{4}(1-\sqrt{x}))}$
Find the area between x-axis and y=f(x) on its maximal domain.

Simpler version:
Show that
$\int_0^1\frac{\sin(\frac{\pi}{2}\sqrt{x})}{2+2\cos(\frac{\pi}{2}\sqrt{x})}-\frac{x+x\tan(\frac{\pi}{4}\sqrt{x})-1}{\tan(\frac{\pi}{4}\sqrt{x})+1}dx\ =\ \frac{2\ln2}{\pi}$

fan96 · Dec 18, 2018

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
Continue to have fun with trig.

Harder version:
$f(x)=\frac{\sin(\frac{\pi}{2}\sqrt{x})}{\sqrt{2}+ \sqrt{2}\cos(\frac{\pi}{2}\sqrt{x})}+\frac{x\sec (\frac{\pi}{4})}{ \tan(\frac{\pi}{4}\sqrt{1-x})+1}-\frac{x\sin(\frac{\pi}{4}\sqrt{x})}{\cos(\frac{\pi}{4}(1-\sqrt{x}))}$
Find the area between x-axis and y=f(x) on its maximal domain.

Simpler version:
Show that
$\int_0^1\frac{\sin(\frac{\pi}{2}\sqrt{x})}{2+2\cos(\frac{\pi}{2}\sqrt{x})}-\frac{x+x\tan(\frac{\pi}{4}\sqrt{x})-1}{\tan(\frac{\pi}{4}\sqrt{x})+1}dx\ =\ \frac{2\ln2}{\pi}$

a very nice integral.

stupid_girl · Dec 20, 2018

Re: HSC 2018 MX2 Integration Marathon

This one should be considerably easier than the previous one.

$\int_{-1}^{1}\frac{x^{2018}\sqrt{1-\cos^2(\frac{\pi}{2}x^{2019})} \log_2(\sec(\frac{\pi}{4}x^{2019}))}{(3+\cos(\pi x^{2019}))(1+2018^x)}dx$

The answer is pretty small. (1/32304)

fan96 · Dec 23, 2018

Re: HSC 2018 MX2 Integration Marathon

I've reduced the integral to

$\frac{1}{4038 \pi \log 2} \int_{0}^1 \frac{\tan^{-1}x}{1+x}\,dx$

if someone else wants to finish it from this, but it seems very difficult.

Maybe a different approach might be necessary?

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