Math Ans=-cot^2x+c/2?

∫cosec³(x) cos(x) dx

= ∫cosec³(x) [cos(x) dx]

= ∫[sin(x)]⁻³ (d sinx)

= (-1/2) [sin(x)]⁻² + C₁

= (1/2) [-cosec²(x)] + (1/2)C

= [-cosec²(x) + C] / 2

Hello,

∫ csc³x cosx dx =

let's split csc³x into csc²x cscx:

∫ csc²x cscx cosx dx =

let's rewrite cscx as 1 /sinx:

∫ csc²x (1 /sinx) cosx dx =

∫ csc²x (cosx /sinx) dx =

∫ csc²x cotx dx =

∫ cotx csc²x dx =

let:

cotx = u

(differentiating both sides)

d(cotx) = du

- csc²x dx = du

csc²x dx = - du

yielding, by substitution:

∫ cotx csc²x dx = ∫ u (- du) =

- ∫ u du =

- [1/(1+1)] u¹ ⁺ ¹ + C =

- (1/2)u² + C

let's substitute back cotx for u, ending with:


- (1/2)cot²x + C



I hope it helps