Question

What is the Integral of `((sec x)^3/tan x) dx`?

Answer 1

`I=ln|cscx-cotx|+secx+C`

Explanation:

Here,

`I=int(secx)^3/tanx dx`

`=int(secx(color(red)(sec^2x)))/tanx dx`

`=int(secx(color(red)(1+tan^2x)))/tanx dx`

`=int(secx+secxtan^2x)/tanx dx`

`=intsecx/tanxdx+int(secxtan^2x)/tanx dx`

`=int(1/cosx)/(sinx/cosx) dx+intsecxtanx dx`

`=intcscxdx+intsecxtanxdx`

`=ln|cscx-cotx|+secx+C`

Answer 2

`1/2ln|(cosx-1)/(cosx+1)|+1/cosx+C, or,`,

`lnsqrt|((cosx-1)/(cosx+1))|+secx+C, or,`,

`ln|tan(x/2)|+secx+C`.

Explanation:

Suppose that, `I=intsec^3x/tanxdx`.

`:. I=int1/cos^3x*cosx/sinxdx=int1/(cos^2xsinx)dx`.

`=intsinx/(cos^2x*sin^2x)dx`,

`=intsinx/{cos^2x(1-cos^2x)}dx`,

Letting `cosx=t," so that, "-sinxdx=dt`, we have,

`I=int(-1)/(t^2(1-t^2))dt=int1/{t^2(t^2-1)}dt`,

`=int{(t^2-(t^2-1)}/{t^2(t^2-1)}dt`,

`=int{t^2/{t^2(t^2-1)}-(t^2-1)/{t^2(t^2-1)}}dt`,

`=int{1/(t^2-1)-1/t^2}dt`,

`={1/2ln|(t-1)/(t+1)|-(t^(-2+1)/(-2+1)}`,

`=1/2ln|(t-1)/(t+1)|+1/t`.

Since, `t=cosx`, we have,

`I=1/2ln|(cosx-1)/(cosx+1)|+1/cosx`,

`=lnsqrt|((cosx-1)/(cosx+1))|+secx`,

`rArr I=ln|tan(x/2)|+secx+C`.