MX2 Integration Marathon (5 Viewers)

Drdusk · Aug 2, 2019

blyatman said:
I like how the thread originally started with "Post integration questions within scope of MX2", but now it's just whatever goes haha.

Especially stupid_girls integrals. My friend and I tried the last one but couldn't get it out.

@stupid_girl does the last integral require the use of $\hspace{2mm}sin^{-1} x + cos^{-1} x = \frac{\pi}{2}$ ?

fan96 · Aug 2, 2019

I have a feeling this is not the fastest method...

$I = \int_0^1\left( \arcsin\left(\frac{x}{x+1}\right)\right)^2\,dx$

$=\int_0^{\pi/6} \frac{y^2\cos y}{(1-\sin y)^2}\,dy \quad \left(y = \arcsin \left(\frac{x}{x+1}\right) \iff x = \frac{\sin y}{1-\sin y }\right)$

$=\frac{\pi^2}{18}- 2 \int_0^{\pi/6} \frac{y}{1-\sin y}\,dy \quad \left({\rm IBP:}\,u = y^2, \, dv = \frac{\cos y}{(1-\sin y)^2}\, dy\right)$

$\text{Now,}\, \int \frac{1}{1-\sin y}\,dy = \frac{1}{\tan(y/2)-1} + C.$

(the proof is left as an exercise to the reader.)

$\implies I =\frac{\pi^2}{18} - 2\left(-\frac{\pi/3}{(\tan(\pi/12) - 1)} +2 \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy\right) \quad \left({\rm IBP:}\,u = y, \, dv = \frac{1}{1-\sin y}\, dy\right).$

From here, I will use the result $\tan \pi/12 = 2 - \sqrt 3$ , which is not difficult to verify via the appropriate double angle identity.

$\int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy$

$= \int_0^{2 - \sqrt 3} \frac{2}{(t-1)(1+t^2)}\, dt \quad \left(t = \tan \frac y2\right)$

$= \int_0^{2 - \sqrt 3} \, -\frac{t}{t^2+1}- \frac{1}{t^2+1} + \frac{1}{t-1}\, dt$

$= \log\left( \frac{{|t-1|} }{\sqrt{t^2+1}} \right) - \arctan t \Bigg|_0^{2 - \sqrt 3}$

$=-\frac 12 \log 2 - \frac{\pi}{12}.$

Note that the absolute value is important when evaluating the lower bound.

$\implies I = \frac{\pi^2}{18} - 2\left(\frac \pi 6 + \frac{\pi \sqrt 3}{6}\right)+ 4 \left( \frac 12 \log 2 + \frac \pi {12}\right)$

$= \frac{\pi^2}{18} - \frac{\pi \sqrt 3}{3}+ 2 \log 2$

$\approx 0.1208.$

aa180 · Aug 2, 2019

My method is similar but slightly different.
$\begin{align*} I &= \int_{0}^{a}f(x)dx \\ &= af(a) - \int_{0}^{f(a)}f^{-1}(x)dx \\ &= 1\cdot\frac{\pi^{2}}{36} - \int_{0}^{\frac{\pi^{2}}{36}}\Big (\frac{1}{1-\sin\sqrt{x}}-1\Big )dx \\ &= \frac{\pi^{2}}{18} - \int_{0}^{\frac{\pi^{2}}{36}}\Big (\sec^{2}\sqrt{x} + \frac{\sin\sqrt{x}}{\cos^{2}\sqrt{x}}\Big )dx \\ &= \frac{\pi^{2}}{18} - 2\int_{0}^{\frac{\pi^{2}}{36}}\underbrace{\frac{1}{2\sqrt{x}}\Big (\sec^{2}\sqrt{x} + \frac{\sin\sqrt{x}}{\cos^{2}\sqrt{x}}\Big )}_{u'}\cdot\underbrace{\sqrt{x}}_{v}dx \\ &= \frac{\pi^{2}}{18} - 2\Big [(\tan\sqrt{x}+\sec\sqrt{x})\sqrt{x}\Big ]_{0}^{\frac{\pi^{2}}{36}} + 2\int_{0}^{\frac{\pi^{2}}{36}}\frac{1}{2\sqrt{x}}\Big (\tan\sqrt{x}+\sec\sqrt{x}\Big )dx \\ & \vdots \\ &= \frac{\pi^{2}}{18} - \frac{\pi}{\sqrt{3}} + 2\ln(2). \end{align*}$

HeroWise · Aug 3, 2019

$1.~~\int_{0}^{\infty}\frac{dx}{(1+x^2)^2}~~~(\mathrm{Ans}:~\frac{\pi}{4})$
$2.~~\int_{0}^{\infty}\frac{x}{(1+x^2)^2}dx~~~(\mathrm{Ans}:~\frac{1}{2})$
$3.~~\int_{0}^{\infty}\frac{dx}{1+x^4}~~~(\mathrm{Ans}:~\frac{\pi\sqrt{2}}{4})$
$4.~~\int_{0}^{\infty}\frac{x^2}{x^6+1}dx~~~(\mathrm{Ans}:~\frac{\pi}{6})$
$5.~~\int_{0}^{\infty}\frac{x^{1/3}}{1+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{\sqrt{6}})$
$6.~~\int_{0}^{2\pi}\frac{\cos^2\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{5\pi}{12})$
$7.~~\int_{0}^{2\pi}\frac{\cos3\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{\pi}{12})$
$8.~~\int_{0}^{\infty}\frac{\cos sx}{k^2+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{2k}e^{-ks})$
$9.~~\int_{0}^{\infty}\frac{\cos x}{x^2+a^2}dx~~~(\mathrm{Ans}:~\frac{\pi e^{-a}}{2a})$
$10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}dx$
$11.~~\int_{0}^{\infty}\frac{cos 4x}{x^4+5x^2+4}dx$
$12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}dx~~~(\mathrm{Ans}:~\frac{\pi}{4e}\sin1)$$

These are Mr Blyatman's question. Ill work them up later today

HeroWise · Aug 3, 2019

For no 2 im getting 1/2. Can someone check? Probs a typo

HeroWise · Aug 3, 2019

3rd one was tooo hard.

Had a peek at it. Got up to sophie germains and gave up

HeroWise · Aug 3, 2019

blyatman said:
Yes I just checked it on wolfram, its 1/2. Answer in the book must've been wrong. Will update post.

What book are these questions from?

HeroWise · Aug 3, 2019

Yeah i tried to factorise it but got too much. Will probably try again later tnight

aa180 · Aug 3, 2019

fan96 · Aug 3, 2019

$4. \int_0^\infty \frac{x^2}{x^6+1}\,dx$

$= \frac 13 \int_{0}^\infty \frac{1}{u^2+1}\,du \quad (u = x^3)$

$= \frac 13 \cdot \frac \pi 2$

$= \frac \pi 6.$

fan96 · Aug 3, 2019

blyatman said:
Nice one, didn't think of that haha, trivial using a smart substitution!

Admittedly, I went through a lengthy process (including looking up a standard integral table for $\text{sech}$ ) and it was only at the end I realised that everything I did could be compressed into one substitution.

Paradoxica · Aug 4, 2019

blyatman said:
$8.~~\int_{0}^{\infty}\frac{\cos ax}{x^2+k^2}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{2k}e^{-ak})$
$10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}\,dx$
$11.~~\int_{0}^{\infty}\frac{\cos4x}{x^4+5x^2+4}\,dx$
$12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{4e}\sin1)$
$13.~~\int_{0}^{\infty}\frac{x^{-k}}{x+1}\,dx,~0<k<1~~~(\mathrm{Ans:}~\frac{\pi}{\sin k\pi})$

These are all beyond the scope of the syllabus.

stupid_girl · Aug 4, 2019

fan96 said:
I have a feeling this is not the fastest method...

$I = \int_0^1\left( \arcsin\left(\frac{x}{x+1}\right)\right)^2\,dx$

$=\int_0^{\pi/6} \frac{y^2\cos y}{(1-\sin y)^2}\,dy \quad \left(y = \arcsin \left(\frac{x}{x+1}\right) \iff x = \frac{\sin y}{1-\sin y }\right)$

$=\frac{\pi^2}{18}- 2 \int_0^{\pi/6} \frac{y}{1-\sin y}\,dy \quad \left({\rm IBP:}\,u = y^2, \, dv = \frac{\cos y}{(1-\sin y)^2}\, dy\right)$

$\text{Now,}\, \int \frac{1}{1-\sin y}\,dy = \frac{1}{\tan(y/2)-1} + C.$

(the proof is left as an exercise to the reader.)

$\implies I =\frac{\pi^2}{18} - 2\left(-\frac{\pi/3}{(\tan(\pi/12) - 1)} +2 \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy\right) \quad \left({\rm IBP:}\,u = y, \, dv = \frac{1}{1-\sin y}\, dy\right).$

From here, I will use the result $\tan \pi/12 = 2 - \sqrt 3$ , which is not difficult to verify via the appropriate double angle identity.

$\int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy$

$= \int_0^{2 - \sqrt 3} \frac{2}{(t-1)(1+t^2)}\, dt \quad \left(t = \tan \frac y2\right)$

$= \int_0^{2 - \sqrt 3} \, -\frac{t}{t^2+1}- \frac{1}{t^2+1} + \frac{1}{t-1}\, dt$

$= \log\left( \frac{{|t-1|} }{\sqrt{t^2+1}} \right) - \arctan t \Bigg|_0^{2 - \sqrt 3}$

$=-\frac 12 \log 2 - \frac{\pi}{12}.$

Note that the absolute value is important when evaluating the lower bound.

$\implies I = \frac{\pi^2}{18} - 2\left(\frac \pi 6 + \frac{\pi \sqrt 3}{6}\right)+ 4 \left( \frac 12 \log 2 + \frac \pi {12}\right)$

$= \frac{\pi^2}{18} - \frac{\pi \sqrt 3}{3}+ 2 \log 2$

$\approx 0.1208.$

My method is similar to yours. DI table may make the integration by parts neater.
$D \qquad I$
$x^2 \qquad \frac{\cos x}{\left(1-\sin x\right)^2}$
$2x\qquad \frac{1}{1-\sin x}$
$2\qquad \frac{\cos x}{1-\sin x}$
$0\qquad -\ln\left(1-\sin x\right)$

$\int_{ }^{ }\frac{x^2\cos x}{\left(1-\sin x\right)^2}dx=\frac{x^2}{1-\sin x}-\frac{2x\cos x}{1-\sin x}-2\ln\left(1-\sin x\right)+C$

stupid_girl · Aug 4, 2019

stupid_girl said:
This one is challenging if you can't think of a useful substitution.
$\int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$

Have anyone figured out?

The trick is:
$\int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$
$=\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{8\left(2+\sqrt{5}\right)}{27}\right)^x+\left(\frac{6\left(3+\sqrt{5}\right)}{27}\right)^x}dx$
$=\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{1+\sqrt{5}}{3}\right)^{3x}+\left(\frac{1+\sqrt{5}}{3}\right)^{2x}}dx$

Clearly, the useful substitution is
$u=\left(\frac{1+\sqrt{5}}{3}\right)^x$

stupid_girl · Aug 4, 2019

stupid_girl said:
This one is relatively routine.
$\int_0^{\frac{\pi}{2}}\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx=\frac{\pi}{72}$

This integral itself should be quite routine. Getting the final answer in terms of pi is slightly tricky.

$\int\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx$
$=\frac{1}{5}\int x\frac{\left(5\cos x+30\sin x\right)}{\left(5\sin x-30\cos x+31\right)^2}dx$
$=-\frac{1}{5}\int x\ d\left(\frac{1}{5\sin x-30\cos x+31}\right)$
$=-\frac{x}{5}\left(\frac{1}{5\sin x-30\cos x+31}\right)+\frac{1}{5}\int_{ }^{ }\left(\frac{1}{5\sin x-30\cos x+31}\right)dx$
$=-\frac{x}{5}\left(\frac{1}{5\sin x-30\cos x+31}\right)+\frac{1}{15}\tan^{-1}\left(\frac{61\tan\frac{x}{2}+5}{6}\right)+c$

To get the final answer, you need to show that:
$\tan^{-1}11-\tan^{-1}\frac{5}{6}=\frac{\pi}{4}$

Paradoxica · Aug 5, 2019

stupid_girl said:
Clearly, the useful substitution is
$u=\left(\frac{1+\sqrt{5}}{3}\right)^x$

intuition said it was φ-like but i didn't want to stay up that late to figure it out;

a good one nonetheless

HeroWise · Aug 5, 2019

stupid_girl said:
Have anyone figured out?

The trick is:
$\int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$
$=\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{8\left(2+\sqrt{5}\right)}{27}\right)^x+\left(\frac{6\left(3+\sqrt{5}\right)}{27}\right)^x}dx$
$=\int\frac{1}{\left(\frac{1+\sqrt{5}}{3}\right)^x+\left(\frac{1+\sqrt{5}}{3}\right)^{3x}+\left(\frac{1+\sqrt{5}}{3}\right)^{2x}}dx$

Clearly, the useful substitution is
$u=\left(\frac{1+\sqrt{5}}{3}\right)^x$

How are you supposed to know phi is a good sub here? I mean looking at the original question alone

DrEuler · Aug 7, 2019

HeroWise said:
How are you supposed to know phi is a good sub here? I mean looking at the original question alone

Inspection.

stupid_girl · Aug 9, 2019

HeroWise said:
How are you supposed to know phi is a good sub here? I mean looking at the original question alone

After dividing top and bottom by 27^x, the integral is in the form 1/(p^x+q^x+r^x). If this integral has an elementary form, then there must be a relationship between p,q and r. After some trial and error, you would get that relationship.

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