Question

How do you evaluate `int cos(z)/sin^7(z) dz` for [pi/2, pi/6]?

Answer 1

`I=-21/2`

Explanation:

`sinz=t => coszdz=dt`
`sin(pi/2)=1, sin(pi/6)=1/2`

`I=int_(pi/2)^(pi/6) cosz/sin^7z dz= int_(1)^(1/2) dt/t^7= -1/(6t^6)|_(1)^(1/2)`

`I=-1/(6*1/2^6)-(-1/(6*1))=-64/6+1/6=-63/6`

`I=-21/2`