Question

find the integral of cot^3xcsc^3xdx

Answer 1

i got -csc^5x/5 + csc^3x/3 + C

can someone pls tell me if im doing it right pls let me know thanks

Answer 2

you want integral of cot^3(x)*csc^3(x) dx right?

Answer 3

is this (cotx)^3*(cscx)^3 dx?

Answer 4

yep

Answer 5

sorry about that should have made it more clear

Answer 6

Take the integral:
integral cot^3(x) csc^3(x) dx
For the integrand cot^3(x) csc^3(x), use the trigonometric identity cot^2(x) = csc^2(x)-1:
= integral cot(x) csc^3(x) (csc^2(x)-1) dx
For the integrand cot(x) csc^3(x) (csc^2(x)-1), substitute u = csc(x) and du = -(cot(x) csc(x)) dx:
= - integral u^2 (u^2-1) du
Expanding the integrand u^2 (u^2-1) gives u^4-u^2:
= - integral (u^4-u^2) du
Integrate the sum term by term and factor out constants:
= integral u^2 du- integral u^4 du
The integral of u^4 is u^5/5:
= integral u^2 du-u^5/5
The integral of u^2 is u^3/3:
= u^3/3-u^5/5+constant
Substitute back for u = csc(x):
= (csc^3(x))/3-(csc^5(x))/5+constant
Which is equal to:
Answer: |
| = -1/30 ((5 cos(2 x)+1) csc^5(x))+constant

Answer 7

so it's just a simple substitution