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Integration of cosec^2x

Hi i no from formula sheets that the integral of cosec^2x equals -cotx but i still cant work out how to prove this integral.

so far iv tried a few different things

1. i no that cosec^2=1/sin^2x but this doesnt help much

2. cosec^2x=cot^2x + 1, again... couldn't get much further with that

any help would be greatly appreciated
thanks :smile:
Reply 1
Avatar for yusufu
yusufu
16
differentiate -cotx and voila!
Reply 2
Avatar for Mattios88
Mattios88
OP
thanx that was really helpful, do u no a method of integrating it without using differentiation i.e. going straight from cosec^2x to -cotx
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=∫1u2 du=−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)
e-unit
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=∫1u2 du=−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)


nice one. :top:
Mattios88
Hi i no from formula sheets that the integral of cosec^2x equals -cotx but i still cant work out how to prove this integral.

so far iv tried a few different things

1. i no that cosec^2=1/sin^2x but this doesnt help much

2. cosec^2x=cot^2x + 1, again... couldn't get much further with that

any help would be greatly appreciated
thanks :smile:


Hey, I thought I'd help you out as I've just done this problem myself. Use translation: as sin(x)=cos((pi/2)-x) cosec(x)=sec((pi/2)-x) so cosec^2(x)=sec^2((pi/2)-x)
this is easily integratable as I(sec^2(x))=tanx:
I(sec^2((pi/2)-x))=-tan(pi/2-x)+k = -cot(x)+k
korobeiniki
Hey, I thought I'd help you out as I've just done this problem myself. Use translation: as sin(x)=cos((pi/2)-x) cosec(x)=sec((pi/2)-x) so cosec^2(x)=sec^2((pi/2)-x)
this is easily integratable as I(sec^2(x))=tanx:
I(sec^2((pi/2)-x))=-tan(pi/2-x)+k = -cot(x)+k


I doubt that dude will ever come back on here again lol.

BUT I might as well ask you to note that his question asked how to get the integral of cosec2(x) cosec^2(x) without having prior knowledge that -cot(x) differentiates to it. In your explanation, you use a change of trig functions which is fine but the one part that you haven't explained is how ∫sec2(x) dx=tan(x) \displaystyle \int sec^2(x) \ dx = tan(x) . Of course, we know that if we differentiate tan(x), we get sec^2(x) but this does not solve his original question.

If you want to look at how you could, have a look at t-substitutions.
Reply 7
Avatar for Babblebey
Babblebey
Awesome! This helps a lot. Thanks

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