Share

# Find ∫((3sinθ−2)cosθ)/(5−cos2θ−4sinθ)dθ - CBSE (Science) Class 12 - Mathematics

ConceptMethods of Integration - Integration by Substitution

#### Question

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

#### Solution

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 t24t+4=p

(2t4) dt=dp

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

=int(dp)/p

=log |p|+C1

=log |t24t+4|+C1   ......(2)

and

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

=intdt/(t-2)^2

= ∫(t2)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

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution Find ∫((3sinθ−2)cosθ)/(5−cos2θ−4sinθ)dθ Concept: Methods of Integration - Integration by Substitution.
S