Question

How do you evaluate `\int \frac { \cos \theta } { \sin ^ { 4} \theta } d \theta`?

Answer 1

`-1/(3sin^3x) + c`

Explanation:

This intergal can be computed using an appropriate u substitution

Letting `u` = `sintheta`
then `du = costheta d theta`

Hence `int costheta / (sin^4theta) d theta` becomes `int (du)/u^4` = `int u^-4 du`

`therefore -1/3 u^-3 + c`

Hence via considering `u = sintheta`

The antiderivative is `-1/(3sin^3theta) + c`

Answer 2

`=(-sin^(-3) theta)/3 +C`

Explanation:

Let `" "color (brown)(u= sin theta " "`then`" "du=cos theta d theta`
`" "`
Let us substitute the above equation in the given integral
`" "`
`int(cos theta)/(sin^4 theta) d theta`
`" "`
`=int (du)/u^4`
`" "`
`=intu^(-4) du`
`" "`
`=u^(-4+1)/(-4+1) +C" "` where `" "C in RR`
`" "`
`=u^(-3)/-3 +C`
`" "`
`=(color(brown)(-sin^(-3) theta))/3 +C`