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 = `A d/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 = 4

A = –2

⇒ `B - 4 xx 3/2 = -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 + 4int dt/(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`

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

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

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

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

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