# Thread: HSC 2017 MX2 Integration Marathon (archive)

1. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Another method is as follows (we can replace 2016 with any non-negative even integer N and \pi in the integrand with any positive real constant a, so we will do this).$

$\noindent Let I:= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2+3a^{x}\right)\sin^{N}(x) +\left(4 +5a^x \right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. Then we have$

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2a^{x}+3\right)\sin^{N}(x) +\left(4a^{x} +5\right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. \quad (\star)$

$\noindent So addition yields:$

\begin{align*}2I &= 5\times \frac{\pi}{2} + 9\times \frac{\pi}{2}\quad (\star \star) \\ &= 7\pi \\ \Rightarrow I &= \frac{7\pi}{2}.\end{align*}

$\noindent Justification of steps is left as an exercise for the reader, but some hints are given in the next post.$
$\noindent Equation (\star) follows from using \int_{-b}^{b}f(x)\,\mathrm{d}x = \int_{-b}^{b}f(-x)\,\mathrm{d}x.$

$\noindent To help show (\star \star), recall or prove that \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan^{\alpha}x}\,\mathrm{d}x = \frac{\pi}{4} for any real constant \alpha.$

2. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Another method is as follows (we can replace 2016 with any non-negative even integer N and \pi in the integrand with any non-negative real constant a, so we will do this).$

$\noindent Let I:= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2+3a^{x}\right)\sin^{N}(x) +\left(4 +5a^x \right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. Then we have$

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2a^{x}+3\right)\sin^{N}(x) +\left(4a^{x} +5\right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. \quad (\star)$

$\noindent So addition yields:$

\begin{align*}2I &= 5\times \frac{\pi}{2} + 9\times \frac{\pi}{2}\quad (\star \star) \\ &= 7\pi \\ \Rightarrow I &= \frac{7\pi}{2}.\end{align*}

$\noindent Justification of steps is left as an exercise for the reader, but some hints are given in the next post.$
Ah yes that makes sense, and so much simpler than my proof. Thank you!

3. ## Re: HSC 4U Integration Marathon 2017

This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$

4. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by stupid_girl
This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$
http://imgur.com/a/X7CkW

EDIT: Yeah I have no idea how to make the LaTeX valid here imgur will have to do haha. Also I'm still learning so if I can improve anywhere just tell me what to do.

\begin{align*} \int_{0}^{\frac{\pi^2}{16}} \left(tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}\right)dx \end{align*}

$Let u^2 = x \Rightarrow dx=2udu$

\begin{align*} \therefore \int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}dx = \int_0^{\frac{\pi}{4}}2u\left(tan(u)+\frac{2}{1+ta nu}\right)du \end{align*}

\begin{align*} \int_{0}^{\frac{\pi}{4}}2utanudu&=-2uln(cosu)\Big|_0^\frac{\pi}{4} + 2\int_{0}^{\frac{\pi}{4}}lncosudu\\ &=\tfrac{\pi}{4}ln2 +2\int_{0}^\frac{\pi}{4}lncosudu \end{align*}

\noindent \begin{align*} \int\frac{du}{1+tanu}&=\int\frac{cosudu}{cosu+sinu }\\&=\int\left(\frac{\tfrac{1}{2}cosu+\tfrac{1}{2} sinu}{sinu+cosu}+\frac{\tfrac{1}{2}cosu-\tfrac{1}{2}sinu}{sinu+cosu}\right)du &=\tfrac{1}{2}\left(x+ln(cosu+sinu)\right)+C \end{align*}
\begin{align*} \therefore \int_0^\frac{\pi}{4}\frac{4udu}{1+tanu} &= 2u(u+ln(cosu+sinu))\Big|_0^\frac{\pi}{4}-\int_0^\frac{\pi}{4}2udu -2\int_0^\frac{\pi}{4}ln(cosu+sinu)du\\ &=\frac{\pi}{8}\left(\pi+2ln2\right)-\frac{\pi^2}{16}-2\int_0^\frac{\pi}{4}ln(cos(\frac{\pi}{4}-u)+sin(\frac{\pi}{4}-u))du\\ &=\frac{\pi}{16}\left(\pi+4ln2\right) -2\int_{0}^\frac{\pi}{4}ln\sqrt{2}cosudu\\ &=\frac{\pi}{16}\left(\pi+4ln2\right) -2\int_{0}^\frac{\pi}{4}lncosudu-2\int_{0}^\frac{\pi}{4}ln\sqrt{2}du\\ &=\frac{\pi}{16}\left(\pi\right) -2\int_{0}^\frac{\pi}{4}lncosudu \end{align*}

\begin{align*} \therefore \int_{0}^{\frac{\pi^2}{16}} \left(tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}\right)dx = \frac{\pi}{16}(\pi+4ln(2)) \end{align*}

5. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by calamebe
http://imgur.com/a/X7CkW

EDIT: Yeah I have no idea how to make the LaTeX valid here imgur will have to do haha. Also I'm still learning so if I can improve anywhere just tell me what to do.
Still learning what? LaTeX, or Integration?

6. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
Still learning what? LaTeX, or Integration?
LaTeX

EDIT: Though I guess I can always get better at integration and it doesn't stop at high school

7. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by calamebe
LaTeX
I guess you can post your LaTeX attempts (or do it in the LaTeX Practice Thread) and people may give LaTeX tips then (since then we can see what's wrong with the code, if anything).

8. ## Re: HSC 2017 MX2 Integration Marathon

By considering
$\int_{0}^{\infty} f\left((x-x^{-1})^{2}\right) dx$

evaluate:
$\int_{0}^{\infty} \frac{x^{2n-2}}{(x^4-x^{2}+1)^{n}} dx$

9. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by stupid_girl
This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$
I disagree. With the initial substitution $u^2=x$, the standard reflection formula may be applied and we obtain the following equivalence after simplification:

$I=2 \int_0^\frac{\pi}{4} u \left(\tan{u}+\frac{2}{1+\tan{u}}\right) \text{d}u=2\int_0^\frac{\pi}{4} \left(\frac{\pi}{4}-u\right) \left(\tan{u}+\frac{2}{1+\tan{u}}\right)\text{d}u$

$I=\frac{\pi}{4}\int_0^\frac{\pi}{4} \left(\tan{u}+\frac{2}{1+\tan{u}}\right)\text{d}u$

Split and reflect the integral again, but carefully, as follows:

$\tan{u}+\frac{2}{1+\tan{u}} = \tan{u} + \tan{\left(\frac{\pi}{4}-u \right)} + 1$

The value of the integral now simplifies into:

$I = \frac{\pi}{2}\int_0^\frac{\pi}{4} \tan{u} \, \text{d}u + \frac{\pi^2}{16}=\frac{\pi\log{2}}{2} + \frac{\pi^2}{16}$

10. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by wu345
By considering
$\int_{0}^{\infty} f\left((x-x^{-1})^{2}\right) dx$

evaluate:
$\int_{0}^{\infty} \frac{x^{2n-2}}{(x^4-x^{2}+1)^{n}} dx$
Using the reflection across the real line with u+x=0, and Glasser's Master Theorem (ceebs showing the intended way but KingOfActing can do that), the integral simplifies completely into:

$\int_0^\infty \frac{\text{d}x}{(x^2+1)^n}$

This can easily be evaluated in the general form using Differentiation Under The Integral Sign.

11. ## Re: HSC 2017 MX2 Integration Marathon

$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}} dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4 .This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$

12. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by BenHowe
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}} dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4 .This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$
No 2 is easy:

$= \int \frac{d(e^x +1)}{e^x + 1} = \int \frac {d (apple)}{apple} = ln(apple) + C= ln(e^x + 1) + C$

13. ## Re: HSC 2017 MX2 Integration Marathon

Can't you do two by inspection?

14. ## Re: HSC 2017 MX2 Integration Marathon

Reckon q2 is 2u?

15. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by Rathin
Reckon q2 is 2u?
Perhaps. That's why I can do this one.

16. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by si2136
Can't you do two by inspection?
Yes. I did it by inspection but had to show the steps.

17. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by BenHowe
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}} dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4 .This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$
http://m.imgur.com/j3kDVl4

Q1 answer, it's messy so if you can't understand it or read it then tell me. I don't know if there's a faster method but if there is please tell me.

18. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by pikachu975
http://m.imgur.com/j3kDVl4

Q1 answer, it's messy so if you can't understand it or read it then tell me. I don't know if there's a faster method but if there is please tell me.
Correct! One of those ones where you have to complete the square then do the appropriate trig sub and fiddle with a right angled triangle...

19. ## Re: HSC 2017 MX2 Integration Marathon

Are you sure Q3 has closed form solution?

20. ## Re: HSC 2017 MX2 Integration Marathon

I don't know what closed form is but you just need to be creative

21. ## Re: HSC 2017 MX2 Integration Marathon

For Q4, if you stare at it long enough, you should be able to use reverse quotient rule.

Let y=1+ln x, then y'=1/x
The integrand can be written as
(y-1)/y^2
=(x'y-xy')/y^2.
Now, the answer should be obvious.

22. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by stupid_girl
For Q4, if you stare at it long enough, you should be able to use reverse quotient rule.

Let y=1+ln x, then y'=1/x
The integrand can be written as
(y-1)/y^2
=(x'y-xy')/y^2.
Now, the answer should be obvious.
Also can work by $\times\frac{x}{x}$

23. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by BenHowe
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}} dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4 .This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$
4.
$\int \frac{\ln x}{(1+\ln x)^2} dx \\ = \int \frac{\ln x+1-\frac{x}{x}}{(1+\ln x)^2} dx \\ =\frac{x}{1+\ln x}+ C$

24. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by jathu123
4.
$\int \frac{\ln x}{(1+\ln x)^2} dx \\ = \int \frac{\ln x+1-\frac{x}{x}}{(1+\ln x)^2} dx \\ =\frac{x}{1+\ln x}+ C$
Correct!

Sorry if this seems like a dumb question, how do you go from line 2-3? Cause I can only do it your way by splitting the integral to $\int\frac{1}{1+lnx}dx-\int\frac{x}{x(1+lnx)^{2}}dx$. Then if I do the 2nd integral by IBP the remaining integrals cancel, I get the answer.

25. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by BenHowe
Sorry if this seems like a dumb question, how do you go from line 2-3?
My bad, should have wrote it there. I just used reverse quotient rule, with the denominator being 1+lnx and the numerator x

Ohhh just realized stupid_girl already done this above

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