Integration of cosec^2x Watch

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Mattios88
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#1
Report Thread starter 13 years ago
#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
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yusufu
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#2
Report 13 years ago
#2
differentiate -cotx and voila!
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Mattios88
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Report Thread starter 13 years ago
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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
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meef cheese
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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)
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The Orientalist
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(Original post by e-unit)
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:
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korobeiniki
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(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
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Clarity Incognito
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(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  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  \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.
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Babblebey
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Awesome! This helps a lot. Thanks
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