The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

c4 integration

how would I integrate cos(x) cosec^2(x)
Reply 1
Avatar for Wesbian
Wesbian
1
Try rewriting the expression as
cot(x) cosec(x)


Posted from TSR Mobile
Reply 2
Avatar for james22
james22
16
u=sin(x) should do it.
Original post by Wesbian
Try rewriting the expression as
cot(x) cosec(x)


Posted from TSR Mobile


The result becomes pretty much obvious after that. Peace.
Reply 4
Avatar for ztibor
ztibor
10
Original post by Shady778
how would I integrate cos(x) cosec^2(x)


by recognition

∫cos⁡x⋅(sin⁡x)−2dx\displaystyle \int \cos x\cdot \left (\sin x\right )^{-2} dx
Reply 5
Avatar for Shady778
Shady778
OP
11
Original post by ztibor
by recognition

∫cos⁡x⋅(sin⁡x)−2dx\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.
Reply 6
Avatar for Wesbian
Wesbian
1
It's not -cosec(x) it's just cosec(x)

Edit: ignore this

Posted from TSR Mobile
(edited 10 years ago)
Reply 7
Avatar for ztibor
ztibor
10
Original post by Wesbian
It's not -cosec(x) it's just cosec(x)


Posted from TSR Mobile


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

∫cos⁡x⋅cosec2xdx=∫cos⁡x⋅(sin⁡x)−2dx=\displaystyle \int \cos x \cdot cosec^2 x dx =\int \cos x\cdot \left (\sin x\right )^{-2} dx=
=(sin⁡x)−1−1+C=−1sin⁡x+C\displaystyle =\frac{\left (\sin x\right )^{-1}}{-1}+C=-\frac{1}{\sin x}+C
(edited 10 years ago)
Original post by black gang sn1
...


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


ddx(−csc⁡x)=−ddx(csc⁡x)=−sin⁡xddx(1)−1ddx(sin⁡x)sin⁡2x=−−cos⁡xsin⁡2x=cos⁡xsin⁡x⋅1sin⁡x=csc⁡x⋅cot⁡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}

∴∫csc⁡xcot⁡x dx=−csc⁡x+C\therefore \displaystyle \int \csc x \cot x \ dx = -\csc x + \mathcal{C}
(edited 10 years ago)
Reply 9
Avatar for Wesbian
Wesbian
1
Oh sorry my bad :wink: oops


Posted from TSR Mobile

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you