# Thread: HSC 2018-2019 MX2 Integration Marathon

1. ## Re: HSC 2018 MX2 Integration Marathon

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

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

2. ## 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

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

3. ## 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:

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

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:

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

4. ## Re: HSC 2018 MX2 Integration Marathon

Maybe something a bit easier:

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

5. ## Re: HSC 2018 MX2 Integration Marathon

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

6. ## Re: HSC 2018 MX2 Integration Marathon

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

7. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by HeroWise
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-.

8. ## Re: HSC 2018 MX2 Integration Marathon

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

9. ## 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.

10. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by fan96
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.

11. ## 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
$\int_0^1\frac{\sin(\frac{\pi}{2}\sqrt{x})}{2+2\cos (\frac{\pi}{2}\sqrt{x})}-\frac{x+x\tan(\frac{\pi}{4}\sqrt{x})-1}{\tan(\frac{\pi}{4}\sqrt{x})+1}dx\ =\ \frac{2\ln2}{\pi}$

12. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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.

13. ## 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)

14. ## 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?

15. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by fan96
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?
You are almost there. If you put x=tan theta, does it look familiar? You've solved that in the previous one.

16. ## Re: HSC 2018 MX2 Integration Marathon

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

$=\int_{-1}^{1}\frac{x^{2018}}{1+2018^x} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ (2+2 \cos^2(\frac{\pi}{2} x^{2019})) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx \qquad (*)$

By using

$\int_b^a f(x) \, dx = \int_b^a f(a+b-x) \, dx$

and some manipulation, similar to the previous integral, we get

$I =\int_{-1}^{1}\frac{2018^x x^{2018}}{1+2018^x} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ 2+2 \cos^2(\frac{\pi}{2} x^{2019}) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx \quad (**)$

$(*) + (**) \implies$

$2I =\int_{-1}^{1} x^{2018} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ 2+2 \cos^2(\frac{\pi}{2} x^{2019}) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx$

$I =-\frac{1}{4 \log 2} \cdot \frac{4}{2019\pi}\int_{- \pi/ 4}^{\pi/ 4} \frac{|\sin 2\theta| }{ 1 + \cos^2 2\theta } \cdot {\log (\cos \theta)}\, d\theta \qquad \left(\theta = \frac{\pi}{4}x^{2019}\right)$

Because the integrand is even,

$I =-\frac{2}{2019\pi \log 2} \int_{0}^{\pi/ 4} \frac{\sin 2\theta }{ 1 + \cos^2 2\theta } \cdot {\log (\cos \theta)}\, d\theta$

$=-\frac{2}{2019\pi \log 2} \int_{0}^{\pi/ 4} \frac{2 \sin \theta \cos \theta }{ 1 + (2 \cos^2 \theta - 1)^2 } \cdot {\log (\cos \theta)}\, d\theta$

$=\frac{4}{2019\pi \log 2} \int_{1}^{1/\sqrt 2} \frac{x}{ 1 + (2x^2 - 1)^2 } \cdot \log x\, dx \qquad (x = \cos \theta)$

Integrating by parts,

$=\frac{1}{2019\pi \log 2} \int_{1/\sqrt 2}^1 \frac {\tan^{-1}(2x^2-1)} x\, dx$

$=\frac{1}{2019\pi \log 2} \int_{1/\sqrt 2}^1 \frac {x\tan^{-1}(2x^2-1)} {x^2}\, dx$

$=\frac{1}{4038\pi \log 2} \int_{0}^1 \frac {\tan^{-1}u} {u+1}\, du \qquad (u = 2x^2-1)$

Because $\tan^{-1} \tan x = x$ for $-\pi/2 < x < \pi/2$,

$=\frac{1}{4038\pi \log 2} \int_{0}^{\pi/ 4}v \cdot \frac {\sec^2 v} {\tan v+1}\, dv \qquad (u = \tan v)$

This integral has been evaluated before to be equal to $\pi/8 \log 2$.

$\therefore I =\frac{1}{32304}$

17. ## Re: HSC 2018 MX2 Integration Marathon

This one is absolutely a beast.
$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx$

18. ## Re: HSC 2018 MX2 Integration Marathon

I have a question, When do you know to use
$\int^a_0f(x)dx=\int^a_0f(a-x)dx$

19. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by HeroWise
I have a question, When do you know to use
$\int^a_0f(x)dx=\int^a_0f(a-x)dx$
when f(x)+f(a-x) is easier to integrate than f(x)

20. ## Re: HSC 2018 MX2 Integration Marathon

This is another beast.
$\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$

21. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
This is another beast.
$\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$
This is a skeleton solution.
By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as
$\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

A tangent substitution will turn it into a format that Wolfram can solve...finally
$\frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
https://www.wolframalpha.com/input/?...+(1-tan%5E2+x)

I know Wolfram used hyperbolic tangent substitution but it is also solvable in MX2 by secant substitution.

Alternatively, if you don't mind handling improper integral, you can do some algebraic manipulation to get:
$\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}-1}{\left(u^{-1}+u\right)\sqrt{\left(u^{-1}+u\right)^2-2}}du+\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}+1}{\left(u^{-1}-u\right)\sqrt{\left(u^{-1}-u\right)^2+2}}du$
Substituting v=u-1+u and w=u-1-u will lead to two improper (but solvable) integrals because u-1 blows up at 0.

22. ## Re: HSC 2018 MX2 Integration Marathon

This one may look simple at the first glance but actually trickier than you may have thought.
I'm sure a lot of people will come up with an answer 2.
$\int_0^{\pi}\sqrt{1+\sin(2x)}dx$

23. ## Re: HSC 2018 MX2 Integration Marathon

Hint:

$\sqrt{x^2} = |x|$

24. ## Re: HSC 2018 MX2 Integration Marathon

I think this is my first "Stupid_girl's Integrals" ill get. Maybe idk

Well didnt get a $2$ but got a $2\sqrt2$ Thats outta be good right?

I had to graph the thing, is it possible without graphing it?

25. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
This is a skeleton solution.
By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as
$\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

A tangent substitution will turn it into a format that Wolfram can solve...finally
$\frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
https://www.wolframalpha.com/input/?...+(1-tan%5E2+x)

I know Wolfram used hyperbolic tangent substitution but it is also solvable in MX2 by secant substitution.

Alternatively, if you don't mind handling improper integral, you can do some algebraic manipulation to get:
$\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}-1}{\left(u^{-1}+u\right)\sqrt{\left(u^{-1}+u\right)^2-2}}du+\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}+1}{\left(u^{-1}-u\right)\sqrt{\left(u^{-1}-u\right)^2+2}}du$
Substituting v=u-1+u and w=u-1-u will lead to two improper (but solvable) integrals because u-1 blows up at 0.
*soluble

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