MX2 Integration Marathon (1 Viewer)

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.

MX2 Integration Marathon (1 Viewer)

Member

Moderator

617 pages

Member

Active Member

Active Member

Active Member

Active Member

Active Member

Member

617 pages

617 pages

-insert title here-

Active Member

Active Member

Active Member

-insert title here-

Active Member

Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)