Find : `int((3sintheta-2)costheta)/(5-cos^2theta-4sintheta)dthetathetadot`

Explanation

We have
`I=int((3sintheta-2)costheta)/({5-cos^(2)theta-4sintheta})d theta.`
`=int((3sin theta - 2 )cos theta)/((4+sin ^(2)theta -4sintheta))d theta `
`=int((3sin theta -2)cos theta )/((sintheta-2)^(2))d theta =int ((3t-2))/((t-2)^(2))dt,`where sin `theta =t.`
`Let ((3t-2))/((t-2)^(2))=(A)/((t-2))+(B)/((t-2)^(2)).`then,
`(3t-2)-=A(t-2)+B.`
Putting `t=2` in (i) ,we get `B=4`.
Comparing the coefficients of t on both sides of (i) , we get `A=3.`
Thus ,`A=3 and B=4.`
`therefore ((3t-2))/((t-2)^(2))=(3)/((t-2))+(4)/((t-2)^(2))`
`implies I=int((3t-2))/((t-2)^(2))dt=int(3)/((t-2))dt+int(4)/((t-2)^(2))dt`
`=3log |t-2|-(4)/((t-2))+C`
`=3log |sintheta-2|-(4)/((sin theta-2))+C`
`=3 log (2-sin theta)+(4)/((2-sin theta))+C [:' (2-sin theta) gt 0].`