Thanks

transform :

x = 2.sin(t)
4 - x² = 4 ( 1 - (x/2)²) = 4.cos²(t)
dx = 2.cos(t).dt

and the problem is now to find an integral for 1/sin(t) :

dt / sin(t) = 2dy / sin(2y) = dy / sin(y).cos(y) = dy / tan(y).cos²(y) = tan'(y).dy / tan(y)

The above solution doesn't make any sense.

x-sub: x = 2sin(theta)
dx =2cos(theta)d(theta)

This becomes integral of 1/(2sin(theta)/sqrt(4-2sin(theta)^2)

This simplifies to 1/2csc(theta)

Take the integral of that.

integral csc(theta)=ln(csc(theta)-cot(theta)+c

Use triangles to find csc(theta) and cot(theta)

csc (theta) = 2/x
cot(theta) = sqrt(4-x^2)/x

becomes
1/2 * ln(2/x - sqrt(4-x^2)/x)+c

This post was edited by thestoryofisaac on Apr 14 2014 02:09am

Where did you get lost ?

Your final line. The initial integral after trig sub is 1/2csc(theta). You did not make this clear.





Where did your second line come from? Is this some sorta identity? I don't understand it. You have to bring back the initial variable used at the start when solving indefinite integrals.

dy / tan(y).cos²(y) is simply = dy/cos(y). you are equating this to dy/sin(y). How are you doing this?

It seems that we agreed up to this point, since csc(theta) = 1/sin(theta) or 1/sin(t). I just left factor 1/2 away (doesn't belong to the real problem).

And to find an indefinite integral of that, I didn't use a formula for csc, but instead, the formula : sin (2y) = 2.sin(y).cos(y)
which is an identity for any real/complex number y.

The derivative of tan(y) with respect to y is 1/cos²(y) (among other possible forms).

That's why (without writing the variable) :

1 / [tan.cos²] = (1/cos²) / tan

which is of the form f ' / f , with f = tan.

Hence, an anti-derivative would be ln |f| + Constant.

Your final result is correct of course, I guess we are not used to manipulate the same trigonometric formulas (hence the different writings).

the half angle ones are pretty standard in usa

https://www.wolframalpha.com/examples/Math.html i used this site for a little guidance to pass cal classes, hopefully it can help, But school only did so much -.-..