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Integration of cosec^2x

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1. 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
2. differentiate -cotx and voila!
3. 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
4. (Original post by e-unit)
nice one.
5. (Original post by 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
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
6. (Original post by 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 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 . 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.
7. Awesome! This helps a lot. Thanks

Updated: November 11, 2014
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