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

Sy123 · Jul 15, 2013

Re: MX2 Integration Marathon

$$Prove$ \ \ \int_0^1 (1-\sqrt{x})^n \ dx = \frac{2}{(n+1)(n+2)}$

Sy123 · Jul 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{dx}{1+ \cos^4 x}$

$= \int \frac{\sec^2 x \ dx}{\cos^2 x + \sec^2 x }$

$u=\tan x$

$=\int \frac{du}{1+u^2 + \frac{1}{1+u^2}}= \frac{u^2+1}{u^4+2u^2+2} \ du$

$Next we find some$ \ \ a, b \ $so that$

$u^4+2u^2+2=(u^2+au+\sqrt{2})(u^2+ bu + \sqrt{2})$

We can find through expansion or equating co-efficients

$a=\sqrt{2\sqrt{2} -2}$

$a=-b$

Now we have:

$\int \frac{u^2+1}{(u^2+at+\sqrt{2})(u^2+ bt + \sqrt{2})} \ du$

Partial fractions, then re-substitution.

asianese · Jul 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$$Prove$ \ \ \int_0^1 (1-\sqrt{x})^n \ dx = \frac{2}{(n+1)(n+2)}$

$u=1-\sqrt x \Rightarrow x=(1-u)^2 \Rightarrow dx = -2(1-u) \mathop{du}$

$I&=\int_1^0 u^n \cdot -2(1-u)\mathop{du} \\ &= 2\int_0^1 u^n(1-u)\mathop{du} \\ & = 2\int_0^1(u^n-u^{n+1})\mathop{du} \\ &=2\left[\frac{u^{n+1}}{n+1}-\frac{u^{n+2}}{n+2}\right]_0^1 \\ & = 2\left[ \frac{1}{n+1}-\frac{1}{n+2}\right] \\ &=2\left[ \frac{n+2-n-1}{(n+1)(n+2)}\right] \\&=\frac{2}{(n+1)(n+2)}.$

$\int \dfrac{\sin x + \cos x}{\sin^2x + \cos^4x} \mathop{\mathrm{d}x}.$

Sy123 · Jul 17, 2013

Re: MX2 Integration Marathon

asianese said:
$u=1-\sqrt x \Rightarrow x=(1-u)^2 \Rightarrow dx = -2(1-u) \mathop{du}$

$I&=\int_1^0 u^n \cdot -2(1-u)\mathop{du} \\ &= 2\int_0^1 u^n(1-u)\mathop{du} \\ & = 2\int_0^1(u^n-u^{n+1})\mathop{du} \\ &=2\left[\frac{u^{n+1}}{n+1}-\frac{u^{n+2}}{n+2}\right]_0^1 \\ & = 2\left[ \frac{1}{n+1}-\frac{1}{n+2}\right] \\ &=2\left[ \frac{n+2-n-1}{(n+1)(n+2)}\right] \\&=\frac{2}{(n+1)(n+2)}.$

$\int \dfrac{\sin x + \cos x}{\sin^2x + \cos^4x} \mathop{\mathrm{d}x}.$

$=\int \frac{\sin x + \cos x}{\sin^2 x + \cos^2x (1-\sin^2 x} \ dx = \int \frac{\sin x + \cos x}{1-\sin^2x \cos^2x} \ dx$

$u=\sin x - \cos x$

$u^2 = 1-2\sin x \cos x$

$\therefore \ \sin x \cos x = \frac{1}{2}(1-u^2)$

$I = \int \frac{du}{1 - \frac{1}{4}(1-u^2)^2} = \int \frac{4 \ du}{4-(1-u^2)^2} = \int \frac{1}{2-(1-u^2)} + \frac{1}{2+ (1-u^2)} \ du$

$= \tan^{-1}u + \frac{1}{2\sqrt{3}} \ln \left (\frac{ \sqrt{3}+u}{\sqrt{3}-u} \right ) + c$

$=\tan^{-1}(\sin x - \cos x) + \frac{1}{2\sqrt{3}} \ln \left(\frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} + \cos x - \sin x} \right ) + \varepsilon$

Sy123 · Jul 19, 2013

Re: MX2 Integration Marathon

$\int_0^1 \frac{x}{1+ \frac{x}{1+ \frac{x}{1+ \frac{x}{1 + \dots}}}} \ dx$

EDIT: Thanks asianese

braintic · Jul 19, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 \frac{x}{1+ \frac{x}{1+ \frac{x}{1+ \frac{x}{1 + \dots}}}} \ dx$

EDIT: Thanks asianese

Is the answer:

$\frac{5\sqrt{5} - 7}{12}$

rolpsy · Jul 19, 2013

Re: MX2 Integration Marathon

$\int^{\pi}_{0} \frac{\mathrm{d}x}{1 + \cos^2(x)}$

Sy123 · Jul 19, 2013

Re: MX2 Integration Marathon

braintic said:
Is the answer:

$\frac{5\sqrt{5} - 7}{12}$

Yep.

rolpsy said:
$\int^{\pi}_{0} \frac{\mathrm{d}x}{1 + \cos^2(x)}$

$First note that$

$\cos^2(x-\pi/2) = \cos^2(x+ \pi / 2)$

$\therefore $for$ \ \ 0\leq x \leq pi \ \ $graph is symetrical about$ \ \ \pi / 2$

$\therefore \ \int_0^{\pi} \frac{1}{1+\cos^2 x} \ dx = 2 \int_0^{\pi /2}\frac{1}{1+\cos^2 x} \ dx$

$\int_0^{\pi /2} \frac{1}{1+\cos^2 x} \ dx = \int_0^{\pi /2} \frac{\sec^2 x}{\sec^2 x + 1} \ dx$

$u=\tan x$

$=\int_0^{\infty} \frac{du}{u^2+2} = \frac{1}{\sqrt{2}} \tan^{-1} (\frac{u}{\sqrt{2}}) |_0^{\infty} = \frac{\pi}{2\sqrt{2}}$

$\therefore \ \int^{\pi}_{0} \frac{\mathrm{d}x}{1 + \cos^2(x)} = \frac{\pi}{\sqrt{2}}$

Sy123 · Jul 19, 2013

Re: MX2 Integration Marathon

$\int \sin^{-1} \sqrt{ \frac{x}{1+x}} \ dx$

twinklegal19 · Jul 19, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sin^{-1} \sqrt{ \frac{x}{1+x}} \ dx$

is the answer

$x\sin^{-1}\sqrt{\frac{x}{1+x}}-\tan^{-1}\sqrt{x}$

?

~~Probably made a mistake since I rushed it. I used integration by parts once, differentiated the inverse sine like hell and everything sort of fell into place.~~

EDIT: NVM, left out an x in one of my earlier lines oops,

twinklegal19 · Jul 19, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sin^{-1} \sqrt{ \frac{x}{1+x}} \ dx$

is it

$x\sin^{-1}\sqrt{\frac{x}{1+x}}-\sqrt{x} +\tan^{-1}\sqrt{x}$

?

edit: just differentiated and it works out, or at least I think it does

Drongoski · Jul 19, 2013

Re: MX2 Integration Marathon

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

$= \int e^{{e^x}+x} dx = \int e^{e^x}e^xdx = \int e^{e^x}de^x = e^{e^x} + C$

Sy123 · Jul 19, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
is it

$x\sin^{-1}\sqrt{\frac{x}{1+x}}-\sqrt{x} +\tan^{-1}\sqrt{x}$

?

edit: just differentiated and it works out, or at least I think it does

Yep that's it.

===

$\int_0^1 x^m(1-x)^n \ dx$

rolpsy · Jul 20, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 x^m(1-x)^n \ dx$

$\\\text{Let } I(m,n) = \int^{1}_{0} x^m(1-x)^n \, \mathrm{d}x \text{ where } n \text{ and } m \text{ are positive integers.}\\\begin{align*}\text{Integrating by parts with } u &= (1-x)^n \hspace{7mm} \text{ and } \hspace{6.5mm} \mathrm{d}v = x^m \, \mathrm{d}x\\\text{so } \quad \mathrm{d}u &= -n(1-x)^{n-1} \, \mathrm{d}x \, \text{ and } v = \frac{1}{m+1}x^{m+1}\end{align*}\\[-10pt]\text{gives }I(m,n) = \frac{n}{m+1} \cdot I(m + 1, n - 1).\\\begin{align*}\text{Hence, } I(m,n) &= \frac{n}{m+1} \cdot \frac{n-1}{m+2} \cdots \frac{1}{m + n} \cdot I(m+n, 0)\\&=\frac{m!\, n!}{(m+n)!}\cdot I(m+n, 0)\\&=\frac{m!\, n!}{(m+n + 1)!} &\text{since } I(m+n, 0) = \frac{1}{m+n+1}.\end{align*}$

cf. beta function

[HR][/HR]
Next:

$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{\cos(x)}{1+\mathrm{e}^x} \, \mathrm{d}x$

asianese · Jul 20, 2013

Re: MX2 Integration Marathon

MX2 students please note that the example Sy123 posted:

$\int_0^1 x^m(1-x)^n \ dx$

is NOT needed for the syllabus. (It is stipulated in the syllabus that this specific example is not included. Other recursive relations involving more than 1 parameter are not needed.)

Nevertheless it is a good practice in recursion.

chris_power_96 · Jul 20, 2013

Re: MX2 Integration Marathon

rolpsy said:
$\\\text [tex]\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{\cos(x)}{1+\mathrm{e}^x} \, \mathrm{d}x$

1?

Sy123 · Jul 20, 2013

Re: MX2 Integration Marathon

rolpsy said:
$\\\text{Let } I(m,n) = \int^{1}_{0} x^m(1-x)^n \, \mathrm{d}x \text{ where } n \text{ and } m \text{ are positive integers.}\\\begin{align*}\text{Integrating by parts with } u &= (1-x)^n \hspace{7mm} \text{ and } \hspace{6.5mm} \mathrm{d}v = x^m \, \mathrm{d}x\\\text{so } \quad \mathrm{d}u &= -n(1-x)^{n-1} \, \mathrm{d}x \, \text{ and } v = \frac{1}{m+1}x^{m+1}\end{align*}\\[-10pt]\text{gives }I(m,n) = \frac{n}{m+1} \cdot I(m + 1, n - 1).\\\begin{align*}\text{Hence, } I(m,n) &= \frac{n}{m+1} \cdot \frac{n-1}{m+2} \cdots \frac{1}{m + n} \cdot I(m+n, 0)\\&=\frac{m!\, n!}{(m+n)!}\cdot I(m+n, 0)\\&=\frac{m!\, n!}{(m+n + 1)!} &\text{since } I(m+n, 0) = \frac{1}{m+n+1}.\end{align*}$

cf. beta function

[HR][/HR]
Next:

$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{\cos(x)}{1+\mathrm{e}^x} \, \mathrm{d}x$

$u=-x$

$I=\int_{-\pi/2}^{\pi /2} \frac{\cos x}{1+e^x} \ dx = \int_{\pi /2}^{-\pi /2} \frac{-\cos(-u)}{1+e^{-u}} \ du= \int_{-\pi /2}^{\pi /2} \frac{\cos u}{1+e^{-u}} \ du$

$\therefore \ 2I = \int_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^{x}} + \frac{\cos x}{1+e^{-x}} \ dx = \int_{-\pi /2}^{\pi /2} \cos x \frac{2+e^{x} +e^{-x}}{2+e^{x} + e^{-x}}$

$=\int_{-\pi/2}^{\pi /2} \cos x \ dx = 2$

$\therefore I = \int_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^x} \ dx = 1$

===============

$\int_{0}^{\pi /2} \sin (2x) \cos(\cos x) \ dx$

chris_power_96 · Jul 20, 2013

Re: MX2 Integration Marathon

Sy123 said:
===============

$\int_{0}^{\pi /2} \sin (2x) \cos(\cos x) \ dx$

1)expand the sin 2x
2)let u=cosx
3)integartion by parts
Answer; 2sin1+2cos1-2?

asianese · Jul 20, 2013

Re: MX2 Integration Marathon

That's correct.

Sy123 · Jul 20, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
1)expand the sin 2x
2)let u=cosx
3)integartion by parts
Answer; 2sin1+2cos1-2?

Yep.

==================================

$\int_2^{1+ \phi} \frac{(x+1)^3 (x-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} \ dx$

$$Where$ \ \phi = \frac{1+ \sqrt{5}}{2}$

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