# Thread: HSC 2017 MX2 Integration Marathon

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

Originally Posted by InteGrand
$\noindent Let t\in \mathbb{R}. Find \int_{0}^{t} e^{-2x}\sin^{2} x\, \mathrm{d}x.$
www.imgur.com/nGjEjX8

Originally Posted by Paradoxica
Let a+b=c where a, b, c are real numbers. Evaluate

$\int \tan{(\text{a}x)}\tan{(\text{b}x)}\tan{(\text{c}x) }\,\text{d}x$
www.imgur.com/O3sZNpH

EDIT: That final c should be swapped with a k, but I can't be bothered to edit it and upload another photo.

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

$Evaluation of \int \frac{x\cos x+1}{\sqrt{2x^3 \cdot e^{\sin x}+x^2}}dx$

$Evaluation of \int^{\infty}_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}dx$

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

I hope this one isn't too bad.

$\int_{0}^{\pi/4} \frac{log_{2}(cos^{sin 2x}x)}{\pi(3+cos4x)}$

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

Originally Posted by stupid_girl
I hope this one isn't too bad.

$\int_{0}^{\pi/4} \frac{log_{2}(cos^{sin 2x}x)}{\pi(3+cos4x)}$
A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?

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

Originally Posted by integral95
A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?
By inspection, you are correct.

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

Originally Posted by juantheron
$Evaluation of \int \frac{x\cos x+1}{\sqrt{2x^3 \cdot e^{\sin x}+x^2}}dx$

$Evaluation of \int^{\infty}_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}dx$
$Use substitution u = 2x \cdot e^{\sin x} +1 and then the substitution w^{2} = u to obtain; \newline
for x \geq 0 : \ln \left( \frac{\sqrt{2x \cdot e^{\sin x} +1}-1}{\sqrt{2x \cdot e^{\sin x} +1}+1} \right) + C \newline
for x < 0 : \ln \left( \frac{1- \sqrt{2x \cdot e^{\sin x} +1}}{1+ \sqrt{2x \cdot e^{\sin x} +1}} \right) + C$

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

Originally Posted by integral95
A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?
It's correct, but I wonder how Paradoxica did it by inspection.

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

This definite integral must be positive because the integrand is a complete square.

(Sorry for the typo in numerator this afternoon.)
$\int_{0}^{\frac{\pi}{4}} \frac{(sin(2x+\frac{\pi}{4})-\frac{1}{\sqrt{2}})^{4}}{4cos^{8}x}$

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

Originally Posted by stupid_girl
This definite integral must be positive because the integrand is a complete square.
$\int_{0}^{\frac{\pi}{4}} \frac{sin(2x+\frac{\pi}{4})-\frac{1}{\sqrt{2}}}{4cos^{8}x}$
61root(2)/840 right? What I did was break up the sin part, get three different integrals, one of which is easy to work out, then for the other two I created a recursion formula for the integral of sec(x)^n from 0 to pi/4, and added/subtracted the three integrals, and got that. I might post working, but it is unbelievably messy.

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

Originally Posted by calamebe
61root(2)/840 right? What I did was break up the sin part, get three different integrals, one of which is easy to work out, then for the other two I created a recursion formula for the integral of sec(x)^n from 0 to pi/4, and added/subtracted the three integrals, and got that. I might post working, but it is unbelievably messy.
Sorry it's my fault. I accidentally missed out the exponent on the numerator when I typed it in latex.

Nevertheless, the mistyped integral is not messy. I just attempted it and the integrand can be broken into 0.5 sin x/(cos x)^7 and -0.5(sec x)^6. Substitue u=cos x and u=tan x respectively.

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

Originally Posted by stupid_girl
Sorry it's my fault. I accidentally missed out the exponent on the numerator when I typed it in latex.

Nevertheless, the mistyped integral is not messy. I just attempted it and the integrand can be broken into 0.5 sin x/(cos x)^7 and -0.5(sec x)^6. Substitue u=cos x and u=tan x respectively.
Oh yeah I meant my working was messy as in illegible haha.

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

Originally Posted by stupid_girl
This definite integral must be positive because the integrand is a complete square.

(Sorry for the typo in numerator this afternoon.)
$\int_{0}^{\frac{\pi}{4}} \frac{(sin(2x+\frac{\pi}{4})-\frac{1}{\sqrt{2}})^{4}}{4cos^{8}x}$
Well this one is 22/7 - pi, I just did some algebra to break down the expression to tan(x)^4(1-tan(x))^4, made a substitution u=tanx and just some relatively easy integration from there. I did have to use the binomial theorem in the last integral though, I feel like there would be a more elegant way to do it.

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

This is of course a proof that 22/7 > pi.

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

Originally Posted by InteGrand
This is of course a proof that 22/7 > pi.
This is obviously false because we all know that pi = 22/7.

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

Originally Posted by Carrotsticks
This is obviously false because we all know that pi = 22/7.
What???? I though pi = 3.14!!! Mind = blown.

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

Originally Posted by Carrotsticks
This is obviously false because we all know that pi = 22/7.
My life has been a lie.

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

Originally Posted by calamebe
Well this one is 22/7 - pi, I just did some algebra to break down the expression to tan(x)^4(1-tan(x))^4, made a substitution u=tanx and just some relatively easy integration from there. I did have to use the binomial theorem in the last integral though, I feel like there would be a more elegant way to do it.
That's correct!!!
I am not aware of any more elegant way to do it.

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

This question relies on Fundamental Theorem of Calculus and substitution.

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

Originally Posted by stupid_girl
This question relies on Fundamental Theorem of Calculus and substitution.
http://imgur.com/a/YGobh

I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.

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

Originally Posted by stupid_girl
This question relies on Fundamental Theorem of Calculus and substitution.
http://imgur.com/a/YGobh

I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.

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

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

I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.
The approach is correct. Well done!

Just a few minor things to point out.
In (a)(i),
f(G(x))g(x)>=f(x)g(x) because g(x)>=0

In (a)(ii),
phi(x)>=0 for all 0
In (b)(ii),
the result of (a)(ii) can be used because 0<=1-g(t)<=1

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

$\int_{-\pi/2}^{\pi/2} \frac{(2+3\pi^x)(sin^{2016} x)+(4+5\pi^x)(cos^{2016} x)}{(1+\pi^x)(sin^{2016} x+cos^{2016} x)}dx$

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

Originally Posted by stupid_girl
$\int_{-\pi/2}^{\pi/2} \frac{(2+3\pi^x)(sin^{2016} x)+(4+5\pi^x)(cos^{2016} x)}{(1+\pi^x)(sin^{2016} x+cos^{2016} x)}dx$
http://imgur.com/a/QAcfP

I feel as though my method is very convoluted, is there a much easier way?

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

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

I feel as though my method is very convoluted, is there a much easier way?
What's your final answer? Can't seem to spot it.

Edit: Oh found it.

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

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

I feel as though my method is very convoluted, is there a much easier way?
$\noindent Another method is as follows (we can replace 2016 with any non-negative even integer N and \pi in the integrand with any positive real constant a, so we will do this).$

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

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

$\noindent So addition yields:$

\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*}

$\noindent Justification of steps is left as an exercise for the reader, but some hints are given in the next post.$

Page 4 of 11 First ... 23456 ... Last

##### Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•