HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

RayRay · Oct 17, 2013

Re: MX2 Integration Marathon

hmmm. i dont understand.

Sy123 · Oct 17, 2013

Re: MX2 Integration Marathon

petermik said:
View attachment 28792

Here is my solution from request

$I = \int_1^{\sqrt{3}} (1+x^2)^{3/2} \ dx$

$x=\tan u$

$I_5 = \int_{\pi/4}^{\pi/3} \sec^5 u \ du = I_3 + \int_{\pi /4}^{\pi /3} \sec^3 u \tan^2 u \ du$

Do that integral by parts using:

$a = \tan u$

$da = \sec^2 u$

$db = \sec^3 u \tan u$

$b = \frac{1}{3} \sec^3 u$

$I_5 = I_3 + \frac{1}{3} \sec^3 u \tan u |_1^{\sqrt{3}} - \frac{1}{3} I_5$

$\frac{4}{3} I_5 = I_3 + \frac{1}{3} (8\sqrt{3} - 2\sqrt{2} )$

$I_3 = \int_{\pi/4}^{\pi/3} \sec^3 u$

$I_3 = \int_{\pi/4}^{\pi/3} \sec u + \int_{\pi /4}^{\pi /3} \sec u \tan^2 u \ du$

And then repeat IBP with a = tan u, db = sec u tan u

ColdLipstick · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Using squeeze theorem$

$$Prove$ \ \ \lim_{n \to \infty} \frac{1}{n} \int_1^2 \left(1 - \frac{1}{x} \right)^n \ dx = 0$

Sy123 · Oct 17, 2013

Re: MX2 Integration Marathon

$\int \sqrt{a+\sqrt{b+\sqrt{x}}} \ dx$

twinklegal19 · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{a+\sqrt{b+\sqrt{x}}} \ dx$

Did this by brute force, probably made a silly somewhere

$u^2=a+\sqrt{b+\sqrt{x}}$

$dx=8[(u^2-a)^2-b](u^2-a)udu$

$8\int u^2[(u^2-a)^2-b](u^2-a)du$

expand, integrate, then sub in

HeroicPandas · Oct 17, 2013

Re: MX2 Integration Marathon

$\int \frac{dx}{e^{\sin\left ( -x \right )}secx}$

twinklegal19 · Oct 17, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$\int \frac{dx}{e^{\sin\left ( -x \right )}secx}$

$=\int e^{-\sin{(-x)}}\cos{x}dx$

$=\int e^{sin{x}}cos{x}dx$

$=e^{\sin{x}}+C$

RealiseNothing · Oct 17, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
$=\int e^{-\sin{x}}\cos{x}dx$

$=\int e^{sin{x}}cos{x}dx$

$=e^{\sin{x}}+C$

Nice first integral.

twinklegal19 · Oct 17, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Nice first integral.

I know right

Sy123 · Oct 17, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
Did this by brute force, probably made a silly somewhere

$u^2=a+\sqrt{b+\sqrt{x}}$

$dx=8[(u^2-a)^2-b](u^2-a)udu$

$8\int u^2[(u^2-a)^2-b](u^2-a)du$

expand, integrate, then sub in

That is essentially it, a better substitution would be, $x= ((u-a)^2 - b)^2$ but it doesn't make too much a difference

HeroicPandas said:
$\int \frac{dx}{e^{\sin\left ( -x \right )}secx}$

$\int \cos x e^{\sin x} \ dx = e^{\sin x} + c$

----

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

HeroicPandas · Oct 17, 2013

Re: MX2 Integration Marathon

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

Rationalise top

$\int \frac{1-x}{\sqrt{1 - x^2}} = \int \frac{1}{\sqrt{1-x^2}} - \int \frac{x}{\sqrt{1-x^2}} = sin^{-1}x + \sqrt{1-x^2} + C$

HeroicPandas · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \cos x e^{\sin x} \ dx = e^{\sin x} + c$
]

twinklegal19 said:
$=\int e^{-\sin{(-x)}}\cos{x}dx$ $=\int e^{sin{x}}cos{x}dx$ $=e^{\sin{x}}+C$

very nice!

Sy123 · Oct 17, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
Rationalise top

$\int \frac{1-x}{\sqrt{1 - x^2}} = \int \frac{1}{\sqrt{1-x^2}} - \int \frac{x}{\sqrt{1-x^2}} = sin^{-1}x + \sqrt{1-x^2} + C$

Yep

---

$\int_{0}^{\pi /2} \frac{1}{1 + (\tan x)^{\pi}} \ dx$

SpiralFlex · Oct 17, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
Rationalise top

$\int \frac{1-x}{\sqrt{1 - x^2}} = \int \frac{1}{\sqrt{1-x^2}} - \int \frac{x}{\sqrt{1-x^2}} = sin^{-1}x + \sqrt{1-x^2} + C$

Or use cos2x substitution

bottleofyarn · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{a+\sqrt{b+\sqrt{x}}} \ dx$

Feel like some math today, brute forced also.

$let\,u=\sqrt {a \sqrt {b \sqrt x}}$
$squaring, x=\frac{u^8}{a^4 b^2}$
$\frac{\mathrm{d} x}{\mathrm{d} u}=\frac{8u^7}{a^4 b^2}$
$then, \int \sqrt{a \sqrt{b \sqrt{x}}}dx = \int \frac{8u^8}{a^4 b^2}du$
$= \frac{8u^9}{9a^4 b^2} + C$
$= \frac{8x \sqrt{a \sqrt {b \sqrt x}}}{9}$

man i'm slow at this latex stuff.

HeroicPandas · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep

---

$\int_{0}^{\pi /2} \frac{1}{1 + (\tan x)^{\pi}} \ dx$

Change tanx into sinx and cosx
$\int \frac{dx}{1+(tanx)^{\pi}} = \int \frac{(cos)^{\pi}}{(sinx)^{\pi}+ (cosx)^{\pi}}dx$

Then prove $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ and use it

$I = \int \frac{sin^{\pi}x}{sin^{\pi}x + cos^{\pi}x}$
Add both, $I = x |_0^{\frac{\pi}{2}} = \frac{\pi}{2}$

sorry for the improper working and bad layout

SpiralFlex · Oct 17, 2013

Re: MX2 Integration Marathon

$1) \int \sqrt{1+2\sqrt{x-x^2}}\;dx$

$2) $Thinking$$

$3) $Thinking (will think of something after counterstrike)$$

HeroicPandas · Oct 17, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
Or use cos2x substitution

ahh, i see ze double angle...

ColdLipstick · Oct 17, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep

---

$\int_{0}^{\pi /2} \frac{1}{1 + (\tan x)^{\pi}} \ dx$

chris_power_96 · Oct 17, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
$1) \int \sqrt{1+2\sqrt{x-x^2}}\;dx$

$2) $Thinking$$

$3) $Thinking (will think of something after counterstrike)$$

1) Use the substitution x=sin^2u
2)(1+2(x(1-x))^.5)^.5 then becomes (1+2cosusinu)^.5
3)We the use the fact that sin^2u+cos^2u=1 to make the above square root into sinu+cosu
4) Then the integral becomes 2sinucos^2u +2sin^2ucosu
5) this becomes (2sin^3u)/3 - (2cos^3u)/3
6) this the becomes (2(x)^1.5)/3 - (2(1-x)^1.5)/3

HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

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