MX2 Integration Marathon (3 Viewers)

stupid_girl · Sep 22, 2019

I regret I have picked these numbers as upper and lower limits. It is quite messy to rationalize the answer.

Nevertheless, it is a good practice for rationalization skills. Isn't it?
$\int_{\frac{3}{8}}^{\frac{5}{12}}\frac{\pi\left(4x^2+1\right)-2}{\left(2x\sin\pi x+\cos\pi x\right)^2}dx=\frac{3759-4525\sqrt{2}+1891\sqrt{3}}{5611}$

fan96 · Sep 22, 2019

$\int_{3/8}^{5/12} \frac{\pi(4x^2+1)-2}{(2x\sin \pi x+\cos \pi x)^2}\,dx$

$= \int_{3/8}^{5/12} \frac{\pi(4x^2+1)-2}{\left(\pm \sqrt{4x^2+1} \sin (\pi x + \arctan(1/2x))\right)^2}\,dx \qquad \text{(auxiliary angle method)}$

$= \int_{3/8}^{5/12} \left(\pi - \frac{2}{4x^2 + 1}\right)\csc^2\left(\pi x + \arctan\frac 1{2x}\right)\,dx$

$= \int_{3\pi/8 + \arctan(4/3)}^{5\pi/12 + \arctan(6/5)} \csc^2 u\,du \qquad \left(u = \pi x + \arctan\frac 1{2x}\right)$

$= \cot u \Big|^{3\pi/8 + \arctan(4/3)}_{5\pi/12+ \arctan(6/5)}$

Applying the cotangent angle sum identity and looking up the exact values of $\tan 5\pi/12$ and $\tan 3\pi/8$ will get rid of all the trig functions from the final answer, but rationalising it... does not look fun.

The auxiliary angle method is quite a neat trick here.

stupid_girl · Sep 23, 2019

This one closely resembles an earlier integral...shouldn't take too long to solve.
$\int_0^{\frac{\pi^2}{4}}\frac{\cos\sqrt{x}+6\sin\sqrt{x}}{\left(5\sin\sqrt{x}-30\cos\sqrt{x}+31\right)^2}dx$

stupid_girl · Sep 24, 2019

A couple more
$\int_0^{\pi^2}\frac{1}{\csc\sqrt{x}+\sin\sqrt{x}}dx=\pi\sqrt{2}\ln(\sqrt{2}+1)$
$\int_{\frac{\pi^2}{16}}^{\frac{9\pi^2}{16}}\frac{\sin\sqrt{x}}{1+\cos^4\sqrt{x}}dx=\frac{\pi\sqrt{2}}{4}(\ln5+2\tan^{-1}2)$

stupid_girl · Sep 28, 2019

TheOnePheeph · Sep 29, 2019

stupid_girl said:
A couple more
$\int_0^{\pi^2}\frac{1}{\csc\sqrt{x}+\sin\sqrt{x}}dx=\pi\sqrt{2}\ln(\sqrt{2}+1)$
$\int_{\frac{\pi^2}{16}}^{\frac{9\pi^2}{16}}\frac{\sin\sqrt{x}}{1+\cos^4\sqrt{x}}dx=\frac{\pi\sqrt{2}}{4}(\ln5+2\tan^{-1}2)$

Again, my scanner cropped a bit of the solutions, pretty much just dus and dxs. These were really good questions though

TheOnePheeph · Sep 29, 2019

stupid_girl said:
$\int_{4}^{9}\left(1+\csc\left(\ln x\right)\right)\sqrt{\frac{\tan\left(\ln\sqrt{x}\right)}{x}}dx=6\sqrt{\tan\left(\ln3\right)}-4\sqrt{\tan\left(\ln2\right)}$

fan96 · Sep 30, 2019

TheOnePheeph · Sep 30, 2019

fan96 said:
$\int_\pi^{3\pi/2} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = -\frac \pi 4$

$I = \int_\pi^{3\pi/2}\frac{\sin x}{\sqrt{1+2\sin x\cos x}}\,dx$
$=\int_\pi^{3\pi/2}\frac{\sin x}{\sqrt{\sin^2 x +2\sin x\cos x + \cos^2 x}}\,dx$
$=\int_\pi^{3\pi/2}\frac{\sin x}{\pm \left( \sin x + \cos x \right)}\,dx$
$\frac{\sin x}{\sqrt{1 + \sin 2x}} < 0 \: \text{for} \: \pi < x < 3\pi/2 \text{, so negative case}$
$I = \int_\pi^{3\pi/2}\frac{\cos x - \sin x - \cos x}{\sin x + \cos x}dx$
$= \ln \left(\sin x + \cos x \right) \Biggr |_{\pi}^{3\pi/2} - \int_\pi^{3\pi/2}\frac{\cos x}{\sin x + \cos x}dx$
$I = - \int_\pi^{3\pi/2}\frac{\cos x}{\sin x + \cos x}dx$
$2I = - \int_\pi^{3\pi/2}\frac{\cos x + \sin x}{\sin x + \cos x}dx$
$= - \int_\pi^{3\pi/2}dx$
$2I = -\frac{\pi}{2}$
$I = -\frac{\pi}{4}$

stupid_girl · Sep 30, 2019

Change the limit to force you to find the primitive function

$\int_{3\pi/2}^{5\pi/4} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = \frac \pi 8 + \frac {\ln2} 4$

sharky564 · Oct 1, 2019

stupid_girl said:
Change the limit to force you to find the primitive function
$\int_{3\pi/2}^{5\pi/4} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = \frac \pi 8 + \frac {\ln2} 4$

Pretty easy by the following:

$\frac{\sin(x)}{\sin(x)+\cos(x)} = \frac{1}{2} \left (1 - \frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)} \right )$

sharky564 · Oct 1, 2019

HeroWise · Oct 1, 2019

Why is everyone abusing kings property

sharky564 · Oct 1, 2019

HeroWise said:
Why is everyone abusing kings property

How so?

stupid_girl · Oct 1, 2019

HeroWise said:
Why is everyone abusing kings property

I don't think this property is abused. For questions like #184, this property can help you avoid IBP. Of course, there are also questions like #183 and #185, where you actually need to get the primitive function by IBP.

Paradoxica · Oct 2, 2019

why is it called king's property anyway

border reflection (or border flip) is the most logical choice of name

regal name has no connection whatsoever

HeroWise · Oct 3, 2019

Only kings can abuse sksk

stupid_girl · Oct 3, 2019

The version that I find most useful is
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

Unfortunately, it can't help me to solve #192.

sharky564 · Oct 4, 2019

stupid_girl said:
The version that I find most useful is
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

Unfortunately, it can't help me to solve #192.

You'll figure it out, I believe in you

stupid_girl · Oct 4, 2019

sharky564 said:
You'll figure it out, I believe in you

Any hint?

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