Shady778
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how would I integrate cos(x) cosec^2(x)
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Wesbian
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Try rewriting the expression as
cot(x) cosec(x)


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james22
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u=sin(x) should do it.
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WhiteGroupMaths
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(Original post by Wesbian)
Try rewriting the expression as
cot(x) cosec(x)


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The result becomes pretty much obvious after that. Peace.
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ztibor
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(Original post by Shady778)
how would I integrate cos(x) cosec^2(x)
by recognition

\displaystyle \int \cos x\cdot \left (\sin x\right )^{-2} dx
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Shady778
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(Original post by ztibor)
by recognition

\displaystyle \int \cos x\cdot \left (\sin x\right )^{-2} dx
Thanks for the reply, my mistake, I was doing maths too late into the night to not figure it out ;( thanks again.
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Wesbian
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It's not -cosec(x) it's just cosec(x)

Edit: ignore this

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ztibor
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(Original post by Wesbian)
It's not -cosec(x) it's just cosec(x)


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No. THere _is_ a minus sign before the cosec
but I prefer using in -1/sin (x) form

\displaystyle \int \cos x  \cdot cosec^2 x dx =\int \cos x\cdot \left (\sin x\right )^{-2} dx=
\displaystyle =\frac{\left (\sin x\right )^{-1}}{-1}+C=-\frac{1}{\sin x}+C
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Khallil
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(Original post by black gang sn1)
...
(Original post by Wesbian)
It's not -cosec(x) it's just cosec(x)
\begin{aligned} \dfrac{\text{d}}{\text{d}x} \left( -\csc x \right) & = - \dfrac{\text{d}}{\text{d}x} \left( \csc x \right) \\ & = -  \dfrac{ \sin x \frac{\text{d}}{\text{d}x} (1) - 1 \frac{\text{d}}{\text{d}x} \left( \sin x \right)}{\sin^2 x}  \\ & = - \dfrac{-\cos x}{\sin^2 x} \\ & = \dfrac{\cos x}{\sin x} \cdot \dfrac{1}{\sin x} \\ & = \csc x \cdot \cot x \end{aligned}

\therefore \displaystyle \int \csc x \cot x \ dx = -\csc x + \mathcal{C}
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Wesbian
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Oh sorry my bad oops


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