1.

Evaluate: `int((3sinx-2)cosx)/(5-cos^2x-4sinx) dx`

Answer» We have
`I=int((3sinx-2)cosx)/({5-(1-sin^(2)x)-4sinx})dx`
`=((3sinx-2)cosx)/({4+sin^(2)x-4sinx})dx=int((3sinx-2)cosx)/((2-sinx)^(2))dx` . . . .(i)
Putting 2-sin x=t, we get sin x=2-t and cos x dx =-dt.
`:." "I=-int({3(2-t)-2})/(t^(2))dt=-int((4-3t))/(t^(2))dt=int((3t-4))/(t^(2))dt`
`=int((3)/(t)-(4)/(t^(2)))dt=3log|t|+(4)/(t)+C`
`=3log|{:(2-sinx):}|+(4)/((2-sinx))+C`.


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