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

SpiralFlex · May 25, 2013

Re: MX2 Integration Marathon

U power 6 substitution is faster

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

jmk123 said:
K I did that and got int(6x^3+6-(1/((x-0.5)^2+0.75)))

Final result is (3x^4)/2+6x-(2/sqrt3).tan^-1((2x-1)/sqrt3)
That right?

The solution I got had ln in it.

SpiralFlex said:
U power 6 substitution is faster

This is what I was aiming at.

Makematics · May 25, 2013

Re: MX2 Integration Marathon

Capt Rifle said:
people utilised integration by parts at my school to solve it so yea..

yeah i guess but it's easier to split the numerator by long division or inspection. could definitely come up in 3u and maybe even 2u

jmk123 · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
The solution I got had ln in it.

This is what I was aiming at.

Could you show me how u did it then? I am wrong haha.

Makematics · May 25, 2013

Re: MX2 Integration Marathon

asianese said:
Alternatively you could have rationalised the numerator and put x^2=sin@.

Alternatively you could let x=tan@ and resolve it to the same thing that sy got

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

jmk123 said:
Could you show me how u did it then? I am wrong haha.

$\int \frac{1}{\sqrt[3]{x} + \sqrt{x}} \ dx$

$u^6=x \ \ 6u^5 \ du = dx$

$\Rightarrow \int \frac{6u^5 \ du }{u^2 + u^3}$

$=6\int \frac{u^3}{u+1} \ du$

$= 6 \int u^2 - \frac{u^2}{u+1} \ du = 6\int u^2 - u + \frac{u}{u+1} \ du = 6\int u^2 - u + 1 - \frac{1}{u+1} \ du$

$= 2u^3 - 3u^2 + 6u - 6\ln(u+1) + c$

$= 2\sqrt{x} - 3\sqrt[3](x) + 6\sqrt[6]{x} - 6\ln(\sqrt[6]{x}+1) + c$

jmk123 · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{1}{\sqrt[3]{x} + \sqrt{x}} \ dx$

$u^6=x \ \ 6u^5 \ du = dx$

$\Rightarrow \int \frac{6u^5 \ du }{u^2 + u^3}$

$=6\int \frac{u^3}{u+1} \ du$

$= 6 \int u^2 - \frac{u^2}{u+1} \ du = 6\int u^2 - u + \frac{u}{u+1} \ du = 6\int u^2 - u + 1 - \frac{1}{u+1} \ du$

$= 2u^3 - 3u^2 + 6u - 6\ln(u+1) + c$

$= 2\sqrt{x} - 3\sqrt[3](x) + 6\sqrt[6]{x} - 6\ln(\sqrt[6]{x}+1) + c$

Ahhh yes I see where I went wrong. Thanks

Sy123 · May 26, 2013

Re: MX2 Integration Marathon

$\int \frac{\cos^{-1}x \tan^{-1}x }{\sqrt{1-x^2}} - \frac{\sin^{-1}x \tan^{-1}x}{\sqrt{1-x^2}} + \frac{\sin^{-1}x \cos^{-1}x }{1+x^2} \ dx$

braintic · May 26, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{\cos^{-1}x \tan^{-1}x }{\sqrt{1-x^2}} - \frac{\sin^{-1}x \tan^{-1}x}{\sqrt{1-x^2}} + \frac{\sin^{-1}x \cos^{-1}x }{1+x^2} \ dx$

From the expanded version of the product rule for differentiating a product of three functions (y=uvw becomes y'=uvw'+uv'w+u'vw), the answer is arcsinx.arccosx.arctanx

(BTW, one of my pet peeves is writing such an integral without brackets.)

Sy123 · Jun 6, 2013

Re: MX2 Integration Marathon

$\int \frac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}} \ dx$

twinklegal19 · Jun 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}} \ dx$

$\int \frac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}}=\int \frac{dx}{\sqrt{x}\left ( 1+x^{\frac{1}{6}} \right )}\\$Let $u^6=x\\6u^5du=dx\\\\\int \frac{6u^5du}{u^3(1+u^2)}\\=6\int\frac{u^2}{1+u^2}\\=6\int\left ( 1-\frac{1}{1+u^2} \right )du\\=6(u-\tan^{-1}u)\\=6(x^{\frac{1}{6}}-\tan^{-1}x^{\frac{1}{6}})+C$

Evaluate

$\int \sqrt{9+x^2}dx$

anomalousdecay · Jun 7, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
Evaluate

$\int \sqrt{9+x^2}dx$

Is the answer:

$x\sqrt{9 + x^2} - x^2sin^{-1}\left( \frac{x}{3}\right)$

?????

asianese · Jun 7, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
Is the answer:

$x\sqrt{9 + x^2} - x^2sin^{-1}\left( \frac{x}{3}\right)$

?????

This isn't correct. You should end up with a secant cubed if you use some substitution.

Makematics · Jun 7, 2013

Re: MX2 Integration Marathon

I did it by using a substitution, but i think it can be done using IBP without a sub (not sure tho)

$$Let$~~ x=3\tan\theta\\\\$Integral resolves to$~~ 9\int\sec^3\theta~d\theta\\\\$Using Integration By Parts, the answer is$\\\\\frac{1}{2}x\sqrt{9+x^2}+\frac{9}{2}\ln \left |x+\sqrt{9+x^2}\right | +c$

anomalousdecay · Jun 7, 2013

Re: MX2 Integration Marathon

I tried Integrating by parts but yeah I was completely wrong.

iBibah · Jun 7, 2013

Re: MX2 Integration Marathon

$\int \sqrt{\tan(x)}dx$

Sy123 · Jun 7, 2013

Re: MX2 Integration Marathon

iBibah said:
$\int \sqrt{\tan(x)}dx$

$u^2 = \tan x$

$2u \ du = (1+u^4) \ dx$

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

$\int \sqrt{\tan x} \ dx = \int \frac{2u^2}{1+u^4} \ du$

$By resolving$

$a^4+1 = (a^2 + a\sqrt{2} + 1)(a^2 - a\sqrt{2} + 1)$

$Then a partial fraction decomposition follows, then the integral is computed easily$

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

$\int_0^1 e^{\sqrt[n]{x}} \ dx$

chris_power_96 · Jun 8, 2013

Re: MX2 Integration Marathon

integrate from pi/2 to 0, 1/(2sin^x + cos^2x),

Sy123 · Jun 8, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
integrate from pi/2 to 0, 1/(2sin^x + cos^2x),

By using s^2+c^2 = 1, then multiplying top and bottom by secant squared

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

$u = \tan x$

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

$= \frac{\sqrt{2}}{2} \frac{\pi}{2} = \frac{\pi}{2\sqrt{2}}$

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

$\int \sqrt[n]{x} \ln x \ dx$

chris_power_96 · Jun 8, 2013

Re: MX2 Integration Marathon

it should be 1/2 instead of 2 in your 4th line

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