# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

$\bg_white \int_0^1 \frac{x e^{x}}{1+ e^{x} } dx$
Are you sure the integrand is correct?

#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

Lets campaign the teachers to give mrbunton the mark!

#### fan96

##### 617 pages
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

##### Member
Re: HSC 2018 MX2 Integration Marathon

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).

##### -insert title here-
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

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

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

$\bg_white And changing limits$

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

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

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

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

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

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

Using the identities:

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

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

the integral reduces to:

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

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

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

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

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

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

Giving:

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

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

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

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

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

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

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

Expanding the square and rationalising the denominator gives

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

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##### -insert title here-
Re: HSC 2018 MX2 Integration Marathon

Continuing on from the simplification

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

Reverse the chain rule twice to obtain:

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

Complete the substitution to obtain:

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

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

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

Which "simplifies" to:

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

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

Maybe something a bit easier:

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

#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

Show that
$\bg_white \int_{0}^{\frac{\pi}{2}} \frac{sin(x+\frac{\pi}{4})}{(2^\pi+16^x)(sin^3x+cos^3x)}dx=\frac{\sqrt{6}\pi}{9(2^\pi)}$
If you don't mind me asking, where do you get these integrals from?

#### fan96

##### 617 pages
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.

#### Attachments

• 120.4 KB Views: 53

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

Continue to have fun with trig.

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

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

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

Continue to have fun with trig.

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

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

#### Attachments

• 168.3 KB Views: 43

#### stupid_girl

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

This one should be considerably easier than the previous one.

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

The answer is pretty small. (1/32304)

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

I've reduced the integral to

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

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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