Question

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

Answer 1

Let I = `int((3sintheta-2)costheta)/(5-cos^2theta-4sin theta)d theta`

`=>I=int((3sintheta-2)costheta)/(5-(1-sin^2theta)-4sintheta)d theta`

`=>I=int((3sintheta-2)costheta)/(sin^2theta-4sin theta+4)d theta`

Now, let sin θ=t.

⇒ cos θ dθ=dt

`:.I=int(3t-2)/(t^2-4t+4)`

`=>3t-2=Ad/dx(t^2-4t+4)+B`

`=>3t-2=A(2t-4)+B`

`=>3t-2=(2A)t+B-4A`

Comparing the coefficients of the like powers of t, we get

`2A=3=>A=3/2`

and

B-4A=-2

`=>B-4xx3/2=-2`

`=>B=-2+6=4`

Substituting the values of A and B, we get

`3t-2=3/2(2t-4)+4`

`:.I=int((3t-2)dt)/(t^2-4t+4)`

`=int((3/2(2t-4)+4)/(t^2-4t+4))dt`

 `=3/2int((2t-4)/(t^2-4t+4))dt+4intdt/(t^2-4t+4)`

 `=3/2I_1+4I_2 `

 Here,

`I_1=int((2t-4)dt)/(t^2-4t+4) `

Now,

`I_2=int((2t-4)dt)/(t^2-4t+4)`

Let t²−4t+4=p

⇒(2t−4) dt=dp

`I_1=int((2t-4)dt)/(t^2-4t+4)`

`=int(dp)/p`

 =log |p|+C₁

=log |t²−4t+4|+C₁   ......(2)

and

`I_2=intdt/(t^2-4t+4)`

`=intdt/(t-2)^2`

= ∫(t−2)⁻² dt

`=(t-2)^(-2+1)/(-2+1)+C_2`

 `=(-1)/(t-2)+C_2 " .......(3)"`

From (1), (2) and (3), we get

`I=3/2log|t^2-4t+4|+4xx-1/(t-2)+C_1+C_2`

`=3/2log|sin^2theta-4sintheta+4|+4/(2-t)+C " (Where C=C1+C2)"`

`=3/2log|(sintheta-2^2)|+4/(2-sin theta)+C`

`=3/2xx2log|sintheta-2|+4/(2-sintheta)+C`

`=3log|2-sintheta|+4/(2-sintheta)+C`