MX2 Integration Marathon (1 Viewer)

HeroWise · Nov 4, 2019

Let t=tanx is a good idea?

stupid_girl · Nov 4, 2019

HeroWise said:
Let t=tanx is a good idea?

Yes, it is a good idea (unless you want to use trig identities many times or prove a reduction formula first)

stupid_girl · Nov 24, 2019

The algebra is quite annoying.

$\int_{0}^{1}\frac{1}{2^{x+1}+\sqrt{4^{x}+2^{x}}+\sqrt{4^{x}-2^{x}}}dx=\frac{6+9\sqrt{2}+6\sqrt{3}-9\sqrt{6}-2\pi}{6\ln2}$

stupid_girl · Nov 24, 2019

HeroWise said:
Let t=tanx is a good idea?

I guess you must have got an answer close to zero.
$\int_{0}^{\frac{\pi}{4}}\left(\tan x-\tan^{2}x\right)^{4}dx$
$=\int_{0}^{1}\frac{\left(t-t^{2}\right)^{4}}{1+t^{2}}dt$
$=\int_{0}^{1}\left(t^{6}-4t^{5}+5t^{4}-4t^{2}+4-\frac{4}{1+t^{2}}\right)dt$
$=\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\tan^{-1}1$
$=\frac{22}{7}-\pi$
$\approx0$

ratman1242 · Nov 26, 2019

This one is fun
$\int_0^{\frac{\pi}{2}} \sin^2 (x) \ln ( \tan (x)) dx$

stupid_girl · Nov 27, 2019

ratman1242 said:
This one is fun
$\int_0^{\frac{\pi}{2}} \sin^2 (x) \ln ( \tan (x)) dx$

With the usual trick, f(x)+f(pi/2-x)=-cos(2x) ln(tan x), which can be evaluated by IBP.

stupid_girl · Nov 27, 2019

$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}\left(f\left(x\right)+f\left(a-x\right)\right)dx$

$\int_{0}^{\frac{\pi}{2}}\sin^{2}\left(x\right)\ln\left(\tan\ x\right)dx$
$=\int_{0}^{\frac{\pi}{4}}\left(\sin^{2}x\ln\left(\tan\ x\right)+\cos^{2}x\ln\left(\cot\ x\right)\right)dx$
$=\int_{0}^{\frac{\pi}{4}}\left(\sin^{2}x\ln\left(\tan\ x\right)-\cos^{2}x\ln\left(\tan\ x\right)\right)dx$
$=-\int_{0}^{\frac{\pi}{4}}\cos2x\ln\left(\tan\ x\right)dx$
$=\left[-\frac{\sin2x}{2}\ln\left(\tan\ x\right)\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}}\frac{\sin2x}{2}\cdot\frac{\sec^{2}x}{\tan x}dx$
$=\left[-\frac{\sin2x}{2}\ln\left(\tan\ x\right)\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}}dx$
$=\left[-\frac{\sin2x}{2}\ln\left(\tan\ x\right)+x\right]_{0}^{\frac{\pi}{4}}$
$=\frac{\pi}{4}$

The following can be solved similarly.
$\int_{0}^{\frac{\pi}{2}}\sin^{4}\left(x\right)\ln\left(\tan\ x\right)dx=\frac{\pi}{4}$

stupid_girl · Nov 27, 2019

This one is absolutely painful.

$\int_{0}^{1}\left(\frac{3e^{x}+1}{e^{x}+\sqrt{x+e^{x}}}+\frac{\left(e^{x}+1\right)^{2}}{x+e^{x}+\sqrt{e^{3x}+xe^{2x}}}\right)dx$
$=\ln\left(e^{2}+e+2e\sqrt{e+1}+1\right)-2+2\sqrt{e+1}-\ln4$

TheOnePheeph · Nov 28, 2019

stupid_girl said:
This one is absolutely painful.
$\int_{0}^{1}\left(\frac{3e^{x}+1}{e^{x}+\sqrt{x+e^{x}}}+\frac{\left(e^{x}+1\right)^{2}}{x+e^{x}+\sqrt{e^{3x}+xe^{2x}}}\right)dx$
$=\ln\left(e^{2}+e+2e\sqrt{e+1}+1\right)-2+2\sqrt{e+1}-\ln4$

HSC is over but a good integral is always fun:

$\text{The given integral is equivalent to:}$

$\int_{0}^{1} \frac{3e^x+1}{e^x+\sqrt{x+e^x}} + \frac{\left(e^x+1\right)^2}{\sqrt{x+e^x}\left(e^x+\sqrt{x+e^x}\right)}dx$

$= \int_{0}^{1} \frac{\left(3e^x+1\right)\sqrt{x+e^x}+\left(e^x+1\right)^2}{\sqrt{x+e^x}\left(e^x+\sqrt{x+e^x}\right)}dx$

$= \int_{0}^{1} \frac{\left(e^x+1+2e^x\right)\sqrt{x+e^x} + e^x\left(e^x+1\right) + \left(e^x+1\right)}{\sqrt{x+e^x}\left(e^x+\sqrt{x+e^x}\right)} dx$

$= \int_{0}^{1} \frac{\left(e^x+1\right) \left(\sqrt{x+e^x}+e^x \right)+\left(e^x+1\right) +2e^x \sqrt{x+e^x}}{\sqrt{x+e^x}\left(e^x+\sqrt{x+e^x}\right)} dx$

$= \int_{0}^{1} \frac{e^x+1}{\sqrt{x+e^x}}dx+2\int_{0}^{1} \frac{\frac{e^x + 1}{2\sqrt{x+e^x}}+ e^x}{e^x+\sqrt{x+e^x}}dx$

$= 2\sqrt{x+e^x}\big|_{0}^{1} + 2\ln{\left(e^x + \sqrt{x+ e^x}\right)}\big|_{0}^{1}$

$= 2\sqrt{e+1}-2+ 2\ln{\left(e+\sqrt{e+1}\right)} -2\ln{2}$

$= \ln{\left(e^2 + 2e\sqrt{e+1} + e + 1\right)} -2 +2\sqrt{e+1} -\ln{4}$

stupid_girl · Nov 28, 2019

stupid_girl said:
The algebra is quite annoying.
$\int_{0}^{1}\frac{1}{2^{x+1}+\sqrt{4^{x}+2^{x}}+\sqrt{4^{x}-2^{x}}}dx=\frac{6+9\sqrt{2}+6\sqrt{3}-9\sqrt{6}-2\pi}{6\ln2}$

I have skipped the steps of rationalization.
$\int_{0}^{1}\frac{1}{2^{x+1}+\sqrt{4^{x}+2^{x}}+\sqrt{4^{x}-2^{x}}}dx$
$=\int_{0}^{1}\frac{2^{-x}}{2+\sqrt{1+2^{-x}}+\sqrt{1-2^{-x}}}dx$
$=-\frac{1}{\ln2}\int_{1}^{\frac{1}{2}}\frac{1}{2+\sqrt{1+u}+\sqrt{1-u}}du$
$=\frac{1}{\ln2}\int_{\frac{1}{2}}^{1}\frac{1}{2+\sqrt{1+u}+\sqrt{1-u}}du$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{2\cos2t}{2+\sqrt{1+\sin2t}+\sqrt{1-\sin2t}}dt$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{2\cos2t}{2+\left|\sin t+\cos t\right|+\left|\sin t-\cos t\right|}dt$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{2\cos2t}{2+2\cos t}dt$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{\cos2t}{1+\cos t}\cdot\frac{1-\cos t}{1-\cos t}dt$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{\left(1-2\sin^{2}t\right)\left(1-\cos t\right)}{\sin^{2}t}dt$
$=\frac{1}{\ln2}\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\left(\csc^{2}t-2-\csc t\cot t+2\cos t\right)dt$
$=\frac{1}{\ln2}\left[-\cot t-2t+\csc t+2\sin t\right]_{\frac{\pi}{12}}^{\frac{\pi}{4}}$
$=\frac{1}{\ln2}\left(-\cot\frac{\pi}{4}-2\cdot\frac{\pi}{4}+\csc\frac{\pi}{4}+2\sin\frac{\pi}{4}+\cot\left(\frac{\pi}{3}-\frac{\pi}{4}\right)+2\cdot\frac{\pi}{12}-\csc\left(\frac{\pi}{3}-\frac{\pi}{4}\right)-2\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\right)$
$=\frac{1}{\ln2}\left(-1-\frac{\pi}{2}+\sqrt{2}+\frac{2}{\sqrt{2}}+\frac{1}{\left(\frac{\sqrt{3}-1}{1+\sqrt{3}\cdot1}\right)}+\frac{\pi}{6}-\frac{1}{\left(\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}-\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\right)}-2\left(\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}-\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\right)\right)$
$=\frac{6+9\sqrt{2}+6\sqrt{3}-9\sqrt{6}-2\pi}{6\ln2}$

stupid_girl · Nov 30, 2019

A new one
$\int_{0}^{\pi^{2}}\frac{1}{1-\sin\sqrt{x}\cos\sqrt{x}}dx=\frac{5\sqrt{3}\pi^{2}}{9}$

Paradoxica · Dec 1, 2019

stupid_girl said:
Taking limit
$\lim_{n\to\infty} \int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{n}+\left(1-\sin\sqrt{x}\right)^{n}}{\left(1+\sin\sqrt{x}\right)^{n+1}+\left(1-\sin\sqrt{x}\right)^{n+1}}dx$
$=\lim_{n\to\infty} 2\pi\int_{0}^{1}\frac{1+t^{2n}}{1+t^{2n+2}}dt$
$=2\pi\int_{0}^{1}\lim_{n\to\infty}\frac{1+t^{2n}}{1+t^{2n+2}}dt$
$=2\pi\int_{0}^{1}dt$
$=2\pi$

Hey you can't just commute the limit with the integral like that. Everyone knows limits don't commute in general.

Please justify the commutation using only techniques from the syllabus.

stupid_girl · Dec 1, 2019

Paradoxica said:
Hey you can't just commute the limit with the integral like that. Everyone knows limits don't commute in general.

Please justify the commutation using only techniques from the syllabus.

It requires uniform convergence. Does the syllabus cover techniques to prove uniform convergence?

HeroWise · Dec 7, 2019

@stupid_girl teach us

stupid_girl · Dec 7, 2019

HeroWise said:
@stupid_girl teach us

It is too abstract. I'll leave it for your university lecturer and tutor.

Drdusk · Dec 7, 2019

stupid_girl said:
It is too abstract. I'll leave it for your university lecturer and tutor.

Is that another way of saying “the proof is left as an exercise for the reader”?

HeroWise · Dec 8, 2019

Yes DrDusk...

stupid_girl · Dec 14, 2019

stupid_girl said:
A new one
$\int_{0}^{\pi^{2}}\frac{1}{1-\sin\sqrt{x}\cos\sqrt{x}}dx=\frac{5\sqrt{3}\pi^{2}}{9}$

This one isn't too hard, right? The same old trick applies.

$\int_{0}^{\pi^{2}}\frac{1}{1-\sin\sqrt{x}\cos\sqrt{x}}dx$
$=\int_{0}^{\pi}\frac{2x}{1-\sin x\cos x}dx$
$=\int_{0}^{\frac{\pi}{2}}\frac{2x}{1-\sin x\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\frac{2x}{1-\sin x\cos x}dx$
$=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{1-\sin x\cos x}dx+\frac{3\pi}{2}\int_{\frac{\pi}{2}}^{\pi}\frac{1}{1-\sin x\cos x}dx$
$=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x}{\sec^{2}x-\tan x}dx+\frac{3\pi}{2}\int_{\frac{\pi}{2}}^{\pi}\frac{\sec^{2}x}{\sec^{2}x-\tan x}dx$
$=\frac{\pi}{2}\int_{0}^{\infty}\frac{1}{t^{2}+1-t}dt+\frac{3\pi}{2}\int_{-\infty}^{0}\frac{1}{t^{2}+1-t}dt$
$=\frac{\pi}{2}\left[\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)\right]_{0}^{\infty}+\frac{3\pi}{2}\left[\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)\right]_{-\infty}^{0}$
$=\frac{\sqrt{3}\pi}{3}\left[\frac{\pi}{2}-\left(-\frac{\pi}{6}\right)\right]+\sqrt{3}\pi\left[-\frac{\pi}{6}-\left(-\frac{\pi}{2}\right)\right]$
$=\frac{5\sqrt{3}\pi^{2}}{9}$

stupid_girl · Dec 16, 2019

$\int_{0}^{\tan^{-1}k}\ln\left(1+k\tan x\right)dx=\frac{\tan^{-1}k}{2}\ln\left(k^{2}+1\right)$
(where k is a real constant)

stupid_girl · Dec 17, 2019

This one is similar but the algebra is slightly longer.
$\frac{d}{dk}\int_{0}^{2\tan^{-1}k}\frac{x}{1+\cos x+k\sin x}dx=\frac{2k^{2}\tan^{-1}k+k\ln\left(k^{2}+1\right)-\left(k^{2}+1\right)\left(\tan^{-1}k\right)\ln\left(k^{2}+1\right)}{k^{2}\left(k^{2}+1\right)}$

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