## Wednesday, May 30, 2007

## Thursday, May 17, 2007

## Wednesday, May 16, 2007

### This is hard! solution

Solution kindly provided by datr. Elaborated a bit by moi.

First let u = 1 - 3x - 2x^2.

Then du/dx = -3 - 4x

So, force the numerator into -3 - 4x, like so:

1 + 2x = -1/2(-2 - 4x) = -1/2(-3 - 4x + 1) = - 1/2(-3 - 4x) - 1/2

Hence, the integral must be split into two integrals, one relatively easy to solve by chain rule since numerator is derivative of the denominator.

First one easily solved using chain rule (or by inspection), and the second one needs a bit more working. Can change the second integral to a standard form if you complete the square and take the factor of 1/sqrt2 out.

Then just use arcsine to finish it off!

First let u = 1 - 3x - 2x^2.

Then du/dx = -3 - 4x

So, force the numerator into -3 - 4x, like so:

1 + 2x = -1/2(-2 - 4x) = -1/2(-3 - 4x + 1) = - 1/2(-3 - 4x) - 1/2

Hence, the integral must be split into two integrals, one relatively easy to solve by chain rule since numerator is derivative of the denominator.

First one easily solved using chain rule (or by inspection), and the second one needs a bit more working. Can change the second integral to a standard form if you complete the square and take the factor of 1/sqrt2 out.

Then just use arcsine to finish it off!

## Monday, May 14, 2007

### This is hard!

Really! This will hopefully be the hardest integration problem I will ever encounter for school!

I won't even bother solving it here using this stupid keyboard, but a tip: hyperbolics. And if you actually bother solving this - then how about writing the working and solution in LaTeX, and I could copy and paste it here so that everyone could enjoy the maths! :)

I won't even bother solving it here using this stupid keyboard, but a tip: hyperbolics. And if you actually bother solving this - then how about writing the working and solution in LaTeX, and I could copy and paste it here so that everyone could enjoy the maths! :)

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