The Student Room Group
inegrating factor e^(int -2cosecx)= e^(-2ln(tan0.5x)0= (tan0.5x)^-2

you should be able to do it from there.

Edit: int (cosecx) is in the formula book and there are two possibilities.
You don't need to multiply it out.

multiply both sides of the equation by the integrating factor to get:

(cosecx+cotx)2dydx2y(cosecx+cotx)2=(cosecx+cotx)2tan(x2)ddx[y(cosecx+cotx)2]=(cosecx+cotx)2tan(x2) (cosecx + cotx)^2 \frac{dy}{dx} - 2y(cosecx + cotx)^2 = (cosecx + cotx)^2 tan(\frac{x}{2}) \\ \frac{d}{dx}[y(cosecx + cotx)^2] = (cosecx + cotx)^2 tan(\frac{x}{2})

then just think of a way to integrate the RHS w.r.t. x

EDIT: or just do it the above way...
Reply 3
Am i suppose to know the half angle formula?

Nevermind, Guess my book never told me the integral of cosec.
Reply 4
I got it in the end.

Hmm, am i missing formula or something?
insparato
I got it in the end.

Hmm, am i missing formula or something?

cosecx  dx=lncosecx+cotx  +  C=lntanx2  +  C \huge \int cosec x \; dx = -ln| cosec x + cot x|\; +\; C = ln|tan \frac{x}{2}| \; +\; C

The second one is in the Edexcel formulae booklet, but not in the Heinemann books.
Reply 6
Thanks, have to know your integration with these.

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