integration of trig (1 Viewer)

[c0okies]

how do i find:

∫ cosec(^3)x dx

 

[Nodice]

Use integration by parts:

Let I = cosec^3(x)

/
|cosec^3(x)dx u=cosec(x) u'=-cot(x).cosec(x) v'=cosec^2(x) v=-cot(x)
/

= -cot(x).cosec(x) - int[cot^2(x).cosec(x)]dx
Changing cot^2(x) into cosec^2(x)-1..
= -cot(x).cosec(x) - int[cosec^3(x)] + int[cosec(x)]
2I = -cot(x).cosec(x) + int[cosec(x)] ----> Integral of cosec(x)= log(cosec(x)-cot(x))
I= 1/2 [-cot(x).cosec(x) + log(cosec[x] - cot[x]) +C

 

[243_robbo]

u=cosec(x) u'=-cot(x).cosec(x) v'=cosec^2(x) v=-cot(x)

how did you find the derivative of cosec (x) and the integral of cosec^2 (x)

 

[243_robbo]

u=cosec(x) u'=-cot(x).cosec(x) v'=cosec^2(x) v=-cot(x)

how did you find the derivative of cosec (x) and the integral of cosec^2 (x)?

ive always been puzzled,

what is the derivative of sec x, and cosec x

 

[shimmerz_777]

Robbo, treat it as function of a function,

secx= (cosx)^-1
d/dx=

-1*-sinx(cosx)^-2

sinx(cosx)^-2 or sinx/(cos^2) x


do the same dor cosec, i think thats the right way of doing it

 

[243_robbo]

i see

 

[c0okies]

thanks nodice

 

[c0okies]

Integral of cosec(x)= log(cosec(x)-cot(x))

the sign should be positive right?

 

[Riviet]

Yep, if you're still unsure, differentiate it and you'll get cosecx.

 

[c0okies]

ok it is ^^