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The 2012 STEP Results Discussion Thread

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Reply 1940

Dirac Spinor

Original post by TheMagicMan

I edited now...forgot to change the limits in my last sub....

Some monstrous work there. Bravo!
This is a STEP question so there must be a quicker way.

Reply 1941

TheMagicMan

Original post by ben-smith

Some monstrous work there. Bravo!
This is a STEP question so there must be a quicker way.

Second hardest integral question I've ever done I reckon...any more where that came from?

(edited 12 years ago)

Reply 1942

Dirac Spinor

Original post by TheMagicMan

Second hardest integral question I've ever done I reckon...any more where that came from?

Not to hand. tbh, do any physics and you'll soon bump into some disgusting integrals.
if that's what you like, I'm sure navier stokes is just round the corner for you.

(edited 12 years ago)

Reply 1943

gff

Original post by ben-smith

...

Which year is this?

Reply 1944

TheMagicMan

Original post by gff

Which year is this?

I'm not sure it was a STEP question...I certainly don't remember doing that inteegral before (and it's too hard for STEP anyway)

Reply 1945

Dog4444

Original post by ben-smith

Anyone want to give me a hint on how to do this integral?

\displaystyle \int_{0}^{\pi/2}\dfrac{9cos^2\theta sin^2\theta}{(cos^3\theta+sin^3 \theta)^2}d\theta

Having one of those days...

May be start like this?

Reply 1946

Dirac Spinor

Original post by gff

Which year is this?

93

omg, this is so annoying:

Spoiler

Reply 1947

Dirac Spinor

Original post by Dog4444

May be start like this?

I've got it now. Thanks for help though. :biggrin:

Reply 1948

TheMagicMan

Original post by ben-smith

93

omg, this is so annoying:

Spoiler

Where does this step come from? But if it's that simple I'm pissed :s-smilie:

(edited 12 years ago)

Reply 1949

Farhan.Hanif93

Original post by TheMagicMan

Where does this step come from?

Symmetry about

\theta=\dfrac{\pi}{4}

(notice that the substitution

\theta = \dfrac{\pi}{2} - \phi

leaves the integral unchanged).

Reply 1950

Dirac Spinor

Original post by TheMagicMan

Where does this step come from? But if it's that simple I'm pissed :s-smilie:

symmetry of curve

Reply 1951

TheMagicMan

Original post by Farhan.Hanif93

Symmetry about

\theta=\dfrac{\pi}{4}

(notice that the substitution

\theta = \dfrac{\pi}{2} - \phi

leaves the integral unchanged).

Yeah okay....I hate when you spend ages finding a really long and complicated solution when there's something that simple right there :angry:

Reply 1952

Farhan.Hanif93

Original post by ben-smith

93

omg, this is so annoying:

Spoiler

I have to say... This infuriates me. There I was, trying to integrate over a contour when I could have just done this.

Reply 1953

Dirac Spinor

Well that was productive...

Reply 1954

Farhan.Hanif93

Original post by TheMagicMan

Yeah okay....I hate when you spend ages finding a really long and complicated solution when there's something that simple right there :angry:

I know the feeling.

Reply 1955

Dog4444

Original post by TheMagicMan

You can turn it into

\frac{-(18 (-1+cos(4 x)))}{(10+6 cos(4 x)+3 sin(2 x)-sin(6 x))}

Can you show some general steps how you got there?

Reply 1956

gff

Original post by ben-smith

Ah, fair enough. I thought about trying to divide by something, but didn't think of using symmetry.

Original post by Farhan.Hanif93

I have to say... This infuriates me. There I was, trying to integrate over a contour when I could have just done this.

I used the quotient rule and turned this into a differential equation, whose solution I reckon you don't want to see. :biggrin:

Reply 1957

TheJ0ker

I'm doing a question and it says show

4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha +\cos\beta

Now I have expressions for both cosa and cos b in terms of the same 2 variables. Is it sufficient for me to just sub into the above equation and show that L.H.S = R.H.S to finish the question?

Reply 1958

Dirac Spinor

Original post by TheJ0ker

I'm doing a question and it says show

4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha +\cos\beta

Now I have expressions for both cosa and cos b in terms of the same 2 variables. Is it sufficient for me to just sub into the above equation and show that L.H.S = R.H.S to finish the question?