HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

Makematics · Jun 28, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^4}{\sqrt{1+x^2}} \ dx$

It is likely that I made a mistake but here goes... Also I skipped a few steps (IBP of sec^3theta dtheta) to save some time.

$\\$Let $x=\tan\theta\\\\dx=\sec^2\theta~d\theta\\\\\sqrt{1+x^2}=\sec\theta\\\\I=\int \frac{x^4}{\sqrt{1+x^2}} \ dx \\\\~~=\int\frac{\tan^4\theta\sec^2\theta}{\sec \theta}~d\theta\\\\~~=\int\tan^3\theta~d(\sec \theta)\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta\sec^2\theta~d\theta\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta(1+\tan^2\theta)d\theta$

$\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta~d\theta-3I\\\\4I=\sec\theta\tan^3\theta-3\int\sec\theta(\sec^2\theta-1)d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\int\sec\theta~d\theta-3\int\sec^3\theta~d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-\frac{3}{2}(\ln{\left | \sec\theta+\tan\theta \right |}+\sec\theta\tan\theta)+c\\\\8I=2\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-3\sec\theta\tan\theta+c\\\\\therefore I=\frac{1}{4}x^3\sqrt{1+x^2}+\frac{3}{8}\ln{\left | x+\sqrt{1+x^2} \right |}-\frac{3}{8}x\sqrt{1+x^2}+c$

braintic · Jun 29, 2013

Re: MX2 Integration Marathon

Makematics said:
It is likely that I made a mistake but here goes... Also I skipped a few steps (IBP of sec^3theta dtheta) to save some time.

$\\$Let $x=\tan\theta\\\\dx=\sec^2\theta~d\theta\\\\\sqrt{1+x^2}=\sec\theta\\\\I=\int \frac{x^4}{\sqrt{1+x^2}} \ dx \\\\~~=\int\frac{\tan^4\theta\sec^2\theta}{\sec \theta}~d\theta\\\\~~=\int\tan^3\theta~d(\sec \theta)\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta\sec^2\theta~d\theta\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta(1+\tan^2\theta)d\theta$

$\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta~d\theta-3I\\\\4I=\sec\theta\tan^3\theta-3\int\sec\theta(\sec^2\theta-1)d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\int\sec\theta~d\theta-3\int\sec^3\theta~d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-\frac{3}{2}(\ln{\left | \sec\theta+\tan\theta \right |}+\sec\theta\tan\theta)+c\\\\8I=2\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-3\sec\theta\tan\theta+c\\\\\therefore I=\frac{1}{4}x^3\sqrt{1+x^2}+\frac{3}{8}\ln{\left | x+\sqrt{1+x^2} \right |}-\frac{3}{8}x\sqrt{1+x^2}+c$

Correct. But you don't need the absolute value.

anomalousdecay · Jun 29, 2013

Re: MX2 Integration Marathon

braintic said:
Correct. But you don't need the absolute value.

In HSC you don't need it. But absolute value is most correct.

Makematics · Jun 29, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
In HSC you don't need it. But absolute value is most correct.

yeah i'm pretty sure they accept either in the hsc.

braintic · Jun 29, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
In HSC you don't need it. But absolute value is most correct.

No, I mean you don't need the absolute value in THIS question, whether or not the HSC generally requires it. The thing you are trying to find the absolute value of CANNOT be negative.

asianese · Jun 29, 2013

Re: MX2 Integration Marathon

$\int \dfrac{\mathop{\mathrm{d}x}}{\sin \left(x-\frac{\pi}{3}\right)\cos x}$

Sy123 · Jun 29, 2013

Re: MX2 Integration Marathon

Makematics said:
It is likely that I made a mistake but here goes... Also I skipped a few steps (IBP of sec^3theta dtheta) to save some time.

$\\$Let $x=\tan\theta\\\\dx=\sec^2\theta~d\theta\\\\\sqrt{1+x^2}=\sec\theta\\\\I=\int \frac{x^4}{\sqrt{1+x^2}} \ dx \\\\~~=\int\frac{\tan^4\theta\sec^2\theta}{\sec \theta}~d\theta\\\\~~=\int\tan^3\theta~d(\sec \theta)\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta\sec^2\theta~d\theta\\\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta(1+\tan^2\theta)d\theta$

$\\~~=\sec\theta\tan^3\theta-3\int\sec\theta\tan^2\theta~d\theta-3I\\\\4I=\sec\theta\tan^3\theta-3\int\sec\theta(\sec^2\theta-1)d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\int\sec\theta~d\theta-3\int\sec^3\theta~d\theta\\\\~~~~=\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-\frac{3}{2}(\ln{\left | \sec\theta+\tan\theta \right |}+\sec\theta\tan\theta)+c\\\\8I=2\sec\theta\tan^3\theta+3\ln{\left | \sec\theta+\tan\theta \right |}-3\sec\theta\tan\theta+c\\\\\therefore I=\frac{1}{4}x^3\sqrt{1+x^2}+\frac{3}{8}\ln{\left | x+\sqrt{1+x^2} \right |}-\frac{3}{8}x\sqrt{1+x^2}+c$

Yep nice work.

asianese said:
$\int \dfrac{\mathop{\mathrm{d}x}}{\sin \left(x-\frac{\pi}{3}\right)\cos x}$

1. Expand denominator:

$\int \frac{dx}{\frac{1}{2}\sin x \cos x- \frac{\sqrt{3}}{2}\cos^2x}$

2. Change to sin2x and cos2x

$\int \frac{4 \dx}{\sin(2x) -\sqrt{3}(\cos 2x + 1)}$

3. Apply the substitution: $t=\tan x$

$\int \frac{4}{\frac{2t}{1+t^2} - \sqrt{3} \left(\frac{1-t^2}{1+t^2} + 1 \right )} \cdot \frac{1}{1+t^2} \ dt$

4. Algebra:

$\int \frac{2}{t - \sqrt{3}} \ dt$

Then:

$2\ln \left |\tan x - \sqrt{3} \right | + c$

Sy123 · Jun 29, 2013

Re: MX2 Integration Marathon

Is there a USYD Integration Bee?

$\int \sin(2x) \sin(\cos x) \ dx$

Drongoski · Jun 29, 2013

Re: MX2 Integration Marathon

Sy123 said:
Is there a USYD Integration Bee?

$\int \sin(2x) \sin(\cos x) \ dx$

$d(cos(cos x)) = -sin(cos x ) \times (-sin x) dx$

$\therefore \int sin 2x \times sin(cos x) dx = 2 \int sin x cos x \times sin(cos x) dx$

$= 2 \int cos x d cos(cos x)) = 2 cos x \times cos(cos x)) - 2 \int cos(cos x)) d cos x$

$= 2 cos x \times cos(cos x)) - 2 sin (cos x) + C$

Sy123 · Jun 29, 2013

Re: MX2 Integration Marathon

Drongoski said:
$d(cos(cos x)) = -sin(cos x ) \times (-sin x) dx$

$\therefore \int sin 2x \times sin(cos x) dx = 2 \int sin x cos x \times sin(cos x) dx$

$= 2 \int cos x d cos(cos x)) = 2 cos x \times cos(cos x)) - 2 \int cos(cos x)) d cos x$

$= 2 cos x \times cos(cos x)) - 2 sin (cos x) + C$

Yep, nice work.

$\int_0^{\pi/2} \frac{\sin(2013x)}{\sin x} \ dx$

EDIT: Forgot limits.

Makematics · Jun 30, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep nice work.

Was my way the method you had in mind?

Sy123 · Jun 30, 2013

Re: MX2 Integration Marathon

Makematics said:
Was my way the method you had in mind?

At first I did IBP with

u=x^3, dv= x/sqrt(1+x^2)

Then did it again, similar length.

asianese · Jul 1, 2013

Re: MX2 Integration Marathon

I don't think there is a Usyd Integ Bee...

asianese · Jul 1, 2013

Re: MX2 Integration Marathon

$\int \dfrac{9+6\sqrt x + x}{x+4\sqrt x} \mathop{\mathrm{d}x}$

Sy123 · Jul 1, 2013

Re: MX2 Integration Marathon

asianese said:
$\int \dfrac{9+6\sqrt x + x}{x+4\sqrt x} \mathop{\mathrm{d}x}$

$\int \frac{(\sqrt{x}+3)^2}{\sqrt{x}(\sqrt{x}+4)} \ dx$

$u= \sqrt{x} + 4$

$(u-4)^2 = x$

$dx = 2(u-4) \ du$

$\int \frac{(u-1)^2 \cdot 2(u-4)}{(u-4)u} \ du$

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

$=u^2 - 4u + 2\ln u + c$

$= (\sqrt{x}+4)^2 - 4(\sqrt{x}+4) +2\ln(\sqrt{x}+4) + c$

Sy123 · Jul 3, 2013

Re: MX2 Integration Marathon

This is a great integral, a little long and the answer isn't very elegant, but the path towards it is I think.

$\int \frac{dx}{\sin^3 x + \cos^3 x}$

Makematics · Jul 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
This is a great integral, a little long and the answer isn't very elegant, but the path towards it is I think.

$\int \frac{dx}{\sin^3 x + \cos^3 x}$

care to share a solution?

Sy123 · Jul 6, 2013

Re: MX2 Integration Marathon

Makematics said:
care to share a solution?

Ok:

$\int \frac{dx}{\sin^3 x + \cos^3 x} = \int \frac{dx}{(\sin x + \cos x)(1-\sin x \cos x} = \int \frac{\cos x - \sin x}{(\cos^2 x - \sin^2 x)(1-\sin x \cos x)}$

$u = \cos x + \sin x$

$du = \cos x - \sin x$

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

Make a right angled triangle, we get that:

$\cos 2x = \sqrt{1 - (u^2-1)^2} = u\sqrt{2-u^2}$

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

Replacing the bottom trig identities with the appropriate forms for u.

$I = \int \frac{2 \ du}{u(3-u^2)\sqrt{2-u^2}} = \int \frac{2\sqrt{2-u^2}}{u(2-u^2)(3-u^2)} \ du$

$I = \int 2\sqrt{2-u^2} \left( \frac{1/6}{u} + \frac{1/\sqrt{2}}{2-u^2} + \frac{1/\sqrt{3}}{3-u^2} \right ) \ du$

Then expand, the middle term can be integrated into something to do with sine inverse.
The first term and the last expanded can be dealt with using a sine substitution.

Very straightforward from there.

EDIT: On hindsight I think I could greatly shorten the length of this solution, perhaps by splitting the fractions of trig integral straight away.

Makematics · Jul 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
Ok:

$\int \frac{dx}{\sin^3 x + \cos^3 x} = \int \frac{dx}{(\sin x + \cos x)(1-\sin x \cos x} = \int \frac{\cos x - \sin x}{(\cos^2 x - \sin^2 x)(1-\sin x \cos x)}$

$u = \cos x + \sin x$

$du = \cos x - \sin x$

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

Make a right angled triangle, we get that:

$\cos 2x = \sqrt{1 - (u^2-1)^2} = u\sqrt{2-u^2}$

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

Replacing the bottom trig identities with the appropriate forms for u.

$I = \int \frac{2 \ du}{u(3-u^2)\sqrt{2-u^2}} = \int \frac{2\sqrt{2-u^2}}{u(2-u^2)(3-u^2)} \ du$

$I = \int 2\sqrt{2-u^2} \left( \frac{1/6}{u} + \frac{1/\sqrt{2}}{2-u^2} + \frac{1/\sqrt{3}}{3-u^2} \right ) \ du$

Then expand, the middle term can be integrated into something to do with sine inverse.
The first term and the last expanded can be dealt with using a sine substitution.

Very straightforward from there.

EDIT: On hindsight I think I could greatly shorten the length of this solution, perhaps by splitting the fractions of trig integral straight away.

wowzer haha thanks that's pretty cool.

Sy123 · Jul 6, 2013

Re: MX2 Integration Marathon

$For some non-negative integer$ \ \ n \ \ $evaluate$

$\int_0^{1/2} \frac{x^{2n}}{1-x^2} \ dx$

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