The Student Room Group
Reply 1
1+cot^2 x = cosec^2 x
Reply 2
And if you integrate cosec^2x you get -cotx
Reply 3
And if you integrate 1 you get x
how did u integrate cosec^2x???
Reply 5
barry_4_england
how did u integrate cosec^2x???

by recognition. it's in the formula book.

oh yeah, don't forget the + c !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1:rolleyes:
lost so many marks from not putting in the +c.....
barry_4_england
lost so many marks from not putting in the +c.....

That's where you tell yourself all those marks went. :p:
Let  I=cosec2(x)dx=1sin2(x)dxdivide  top  and  bottom  by  cos2(x)I=sec2(x)tan2(x)letu=tan(x)dudx=sec2(x)dx=dusec2(x)I=1u2du=1u=1tan(x)=cot(x)Let\; I=\int cosec^2(x)\,dx\\ =\int \frac{1}{sin^2(x)}\,dx\\ \\ divide\; top\; and\; bottom\; by\; cos^2(x)\\ \\ I=\int \frac{sec^2(x)}{tan^2(x)} \\let u = tan(x)\\ \Rightarrow \frac{du}{dx} = sec^2(x)\\ \Rightarrow dx = \frac{du}{sec^2(x)}\\ I=\int \frac{1}{u^2}\, du\\ = \frac{-1}{u}\\ = \frac{-1}{tan(x)}\\ = -cot(x)

EDIT: +c
Reply 9
sorry chaps, but more problems.

using an identity, integrate tan^3 x and tan ^4 x

i know it is that sodding sec/tan/1 identity, but i am being crap and cant do it.

in anticipation of your depressingly speedy replies, thank you.
Reply 10
tan^3x = tanx(sec^2x-1) = tanxsec^2x-tanx = tanxsecx.secx - tanx

you should know how to integrate the second term, and note how i have laid out the first term.
Reply 11
chewwy
tan^3x = tanx(sec^2x-1) = tanxsec^2x-tanx = tanxsecx.secx - tanx

you should know how to integrate the second term, and note how i have laid out the first term.

You could also lay it out as tanx.sec^2 x
yet again

(tan x + sec x)^-2 dx

any thoughts?
i think it can be done using the t = tan (x/2) substitution ... just checking through my workings, but i think it works. Not a very elegant method though.

Latest