Question

Integrate the following w.r.t.x : sec⁴x cosec²x

Answer 1

Let I = `int sec^4x  "cosec"^2x*dx`

= `int sec^4x  "cosec"^2x* sec^2x*dx`

Put tan x = t
∴ sec²x·dx = d
Also, sec²x cosec²x = (1 + tan²x)(1 + cot²x)

= `(1 + t^2)(1 + 1/t^2)`

= `(1 + t^2)((t^2 + 1)/t^2)`

= `(t^4 + 2t^2 + 1)/t^2`

= `t^2 + 2 + (1)/t^2`

∴ I = `int (t^2 + 2 + 1/t^2)*dt`

= `int t^2*dt + 2 int *dt + int 1/t^2*dt`

= `t^3/(3) + 2t + (t^-1)/((-1)) + c`

= `(1)/(3)tan^3x + 2tanx - (1)/tanx + c`

= `(1)/(3cot^3x) + (2)/(cotx) - cot x + c`.