Question

How do you find the indefinite integral of `int csc^2t/cott dt`?

Answer 1

`int csc^2t/cott \ dt = ln |Atant|`

Explanation:

Using the trig identity `1 + cot^2theta -= csc^2theta` we have:

`int csc^2t/cott \ dt = int (1+cot^2t)/cott \ dt`
`" " = int 1/cott+cot^2t/cott \ dt`
`" " = int tant+cott \ dt`

Then using the standard results:

`int tanx \ dt = ln |secx| " "(+c)`
`int cosx \ dt = ln |sinx| " "(+c)`

We get the solution where (`A=`constant):

`int csc^2t/cott \ dt = ln |secx| + ln |sinx| + lnA`
`" "= ln |Asintsect|`
`" "= ln |Asint/cost|`
`" "= ln |Atant|`

Answer 2

`-ln|cott| + C`

Explanation:

Start by rewriting this integral in simpler trig functions (with respect to sine and cosine). We use the identities `csctheta = 1/sintheta` and `cottheta = costheta/sintheta` to accomplish this.

`=> int(1/sin^2t)/(cost/sint) dt`

`=> int(1/sin^2t * sint/cost) dt`

`=> int(1/(sintcost))dt`

`=> int(1/(1/2sin2t))dt`

`=> int(2csc2t)dt`

`=> 2int(csc2t)dt`

We now let `u = 2t`.Then `du = 2dt` and `dt = 1/2du`.

`=> 2int(cscu)1/2du`

`=> int(cscu)du`

This is a trick integral to do. Expand the fraction by `cscu + cotu` to get:

`=>int((csc^2u + cotucscu)/(cscu+ cotu))dt`

Now make `v = cscu + cotu`. This means that `(dv)/(du) = -(csc^2u + cotucscu)`. So:

`=>int((csc^2u + cotucscu)/(v)) * (dv)/(-(csc^2u + cotucscu))`

`=>int(-1/v)dv`

`=> -int(1/v)`

`=> -ln|v| + C`

`=> -ln|cscu + cotu| + C`

`=> -ln|csc(2t) + cot(2t)| + C`

Which can be simplified to `-ln|cott| + C`

Hopefully this helps!