HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

bleakarcher · Aug 5, 2012

Decided to make this thread so we can all be exposed to the more difficult integration problems (and for the fun of it of course).

First 2 questions:

integral of [sqrt(1+sin(2x))] dx

integral of sqrt[tan(x)] dx

Carrotsticks · Aug 5, 2012

Re: MX2 Integration Marathon

Good idea! Everybody loves integration.

JINOUGA · Aug 5, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Good idea! Everybody love's integration.

that apostrophe

for the love of all that is good on earth remove it please

Carrotsticks · Aug 5, 2012

Re: MX2 Integration Marathon

=)

Aesytic · Aug 5, 2012

Re: MX2 Integration Marathon

sqrt(1+sin2x) = sqrt(sin^2(x) + cos^2(x) + 2sinxcosx)
= sqrt(sinx+cosx)^2
=sinx+cosx

.'. integral sqrt(1+sin2x) dx = integral (sinx+cosx) dx
= -cosx + sinx + C

not sure how to do the second one

Shadowdude · Aug 5, 2012

Re: MX2 Integration Marathon

ooh the Actuarial Society had an Integration Bee and they had some real awesome questions...

I might try to acquire that sheet and post a few here. Some of them were very very good.

bloodvial · Aug 5, 2012

Re: MX2 Integration Marathon

bleakarcher said:
integral of sqrt[tan(x)] dx

Hope there's a more efficient way than Wolfram Alpha's... http://www2.wolframalpha.com/Calculate/MSP/MSP40121a2dd6ca6h54fd0800002h0773abf15c356b?MSPStoreType=image/gif&s=20&w=489&h=2660

rolpsy · Aug 5, 2012

Re: MX2 Integration Marathon

$\int ^{\frac{\pi}{2}}_0 \frac{\sin(x)}{1 + \sin^2(x)} \, \mathrm{d}x$

asianese · Aug 5, 2012

Re: MX2 Integration Marathon

$\int_0^{\frac{\pi}{2}} \frac{ \sin{x} dx}{1+\sin^2x} \\= \int_0^{\frac{\pi}{2}} \frac{ \sin x dx}{2-\cos^2x}\\ $Let $ u= \cos x \therefore du=-\sin x dx \\ I=\int_1^0 \frac{-du}{2-u^2}\\ =\int_0^1\frac{du}{2-u^2}\\=\frac{1}{2\sqrt2}\ln\left[\frac{\sqrt2+1}{\sqrt2-1}\right]$

U MAD BRO · Aug 5, 2012

Re: MX2 Integration Marathon

int(exp(-a*r^2+a*r*x*cos(theta))*r))

asianese · Aug 5, 2012

Re: MX2 Integration Marathon

d what?

da? dr? dx? d(costheta)? dtheta?

math man · Aug 5, 2012

Re: MX2 Integration Marathon

for integral of $\sqrt(tanx)$ a good start would be to let $u^2 = tanx$

which reduces the integral to $\int \frac{2u^{2}}{1+u^{4}}du$

which is still challenging. If you want a further hint ask me later or someone else
might give you one

rolpsy · Aug 5, 2012

Re: MX2 Integration Marathon

asianese said:
$\int_0^{\frac{\pi}{2}} \frac{ \sin{x} dx}{1+\sin^2x} \\= \int_0^{\frac{\pi}{2}} \frac{ \sin x dx}{2-\cos^2x}\\ $Let $ u= \cos x \therefore du=-\sin x dx \\ I=\int_1^0 \frac{-du}{2-u^2}\\ =\int_0^1\frac{du}{2-u^2}\\=\frac{1}{2\sqrt2}\ln\left[\frac{\sqrt2+1}{\sqrt2-1}\right]$

good job

. after simplification it becomes:

$\frac{1}{\sqrt{2}} \ln \left (1 + \sqrt{2} \right )$

next:

$\int_0^{\frac{\pi}{2}} \frac{\cos^{2012}(x)}{\sin^{2012}(x) + \cos^{2012}(x)} \, \mathrm{d}x$

edit:

math man said:
for integral of $\sqrt(tanx)$ a good start would be to let $u^2 = tanx$

which reduces the integral to $\int \frac{2u^{2}}{1+u^{4}}du$

which is still challenging. If you want a further hint ask me later or someone else
might give you one

$\\\textup{Roots of } u^4 + 1 \textup{ are } u = \text{cis}\left ( \pm \frac{\pi}{4} \right ), \text{cis}\left ( \pm \frac{3\pi}{4} \right )\\\begin{align*}u^4 + 1 &= \left (u - \text{cis}\left (\frac{\pi}{4} \right ) \right )\left (u - \text{cis}\left (-\frac{\pi}{4} \right ) \right )\left (u - \text{cis}\left (\frac{3\pi}{4} \right ) \right )\left (u - \text{cis}\left (-\frac{3\pi}{4} \right ) \right )\\&=\left ( u^2 -2\cos\left ( \frac{\pi}{4} \right )u + 1 \right )\left ( u^2 -2\cos\left ( \frac{3\pi}{4} \right )u + 1 \right )\\&=\left ( u^2 - u\sqrt{2} + 1 \right )\left ( u^2 + u\sqrt{2} + 1 \right )\end{align*}$

then use partial fractions. very tedious.

U MAD BRO · Aug 5, 2012

Re: MX2 Integration Marathon

$\int (sin7x)^{12}.(cos7x)^3dx$

U MAD BRO · Aug 5, 2012

Re: MX2 Integration Marathon

asianese said:
d what?

da? dr? dx? d(costheta)? dtheta?

This is a double integral over a keystone shape in polar coordinates. (so limits are finite, and not a complete circle).

I've tried all sorts of things. I am not even sure if there is an analytic solution. Is there away to show there isn't one?

rolpsy · Aug 5, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
$\int (sin7x)^{12}.(cos7x)^3dx$

$\begin{align*}\int \sin^{12}\left ( 7x \right )\cos^{3}\left ( 7x \right ) \, \mathrm{d}x &= \int \sin^{12}\left ( 7x \right )\cos\left ( 7x \right ) -\sin^{14}\left ( 7x \right )\cos\left ( 7x \right ) \, \mathrm{d}x\\&=\frac{1}{7\times13}\sin^{13}(7x) - \frac{1}{7\times15}\sin^{15}(7x) + C\end{align*}$

U MAD BRO · Aug 5, 2012

Re: MX2 Integration Marathon

rolpsy said:
$\begin{align*}\int \sin^{12}\left ( 7x \right )\cos^{3}\left ( 7x \right ) \, \mathrm{d}x &= \int \sin^{12}\left ( 7x \right )\cos\left ( 7x \right ) -\sin^{14}\left ( 7x \right )\cos\left ( 7x \right ) \, \mathrm{d}x\\&=\frac{1}{7\times13}\sin^{13}(7x) - \frac{1}{7\times15}\sin^{15}(7x) + C\end{align*}$

all trig integration with powers comes down to getting cos to the power of one =)

HeroicPandas · Aug 5, 2012

Re: MX2 Integration Marathon

Integral of e^[e^(x) +x].dx

U MAD BRO · Aug 5, 2012

Re: MX2 Integration Marathon

$\int e^{e^{x}+x}dx$

HeroicPandas · Aug 5, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
$\int e^{e^{x}+x}dx$

yes do it

its a trick question

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