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3.6 Q-30

Question Statement

Evaluate the following integral:

∫02Sin⁑x(1+Cos⁑x)(2+Cos⁑x),dx\int_{0}^{2} \frac{\operatorname{Sin} x}{(1+\operatorname{Cos} x)(2+\operatorname{Cos} x)} , dx

Background and Explanation

To solve this integral, we use a substitution method. Understanding the following concepts is important:

  • Trigonometric identities: sin⁑x\sin x and cos⁑x\cos x are the core functions involved.
  • Substitution: We’ll make a substitution to simplify the integral.
  • Integration of rational functions: After substitution, the integrand becomes a rational function, which can be split and integrated easily.

Solution

Step-by-step explanation of the solution:

  1. Substitute:
    Let’s make the substitution to simplify the integral. Set:
u=1+Cos⁑x u = 1 + \operatorname{Cos} x

Now, calculate the limits of uu:

  • When x=0x = 0, u=1+cos⁑(0)=2u = 1 + \cos(0) = 2.
  • When x=Ο€2x = \frac{\pi}{2}, u=1+cos⁑(Ο€2)=1u = 1 + \cos(\frac{\pi}{2}) = 1.
  1. Express dudu:
    Differentiate u=1+cos⁑xu = 1 + \cos x with respect to xx:
du=βˆ’sin⁑x,dxorβˆ’du=sin⁑x,dx du = -\sin x , dx \quad \text{or} \quad -du = \sin x , dx
  1. Substitute into the integral:
    Now substitute uu and dudu into the integral:
∫0Ο€/2sin⁑x(1+cos⁑x)(2+cos⁑x),dx=∫21βˆ’duu(1+u+1) \int_{0}^{\pi / 2} \frac{\sin x}{(1 + \cos x)(2 + \cos x)} , dx = \int_{2}^{1} \frac{-du}{u(1+u+1)}

Simplifying the integrand:

=∫21(βˆ’1u+1u+1),du = \int_{2}^{1} \left( \frac{-1}{u} + \frac{1}{u + 1} \right) , du
  1. Separate the integrals:
    Split the integral into two parts:
=βˆ’βˆ«211u,du+∫211u+1,du = -\int_{2}^{1} \frac{1}{u} , du + \int_{2}^{1} \frac{1}{u + 1} , du
  1. Integrate:
    The integrals are simple logarithmic integrals:
=βˆ’ln⁑u∣21+ln⁑(u+1)∣21 = - \ln u \Big|_2^1 + \ln (u + 1) \Big|_2^1
  1. Evaluate the bounds:
    Now evaluate the logarithmic expressions:
=βˆ’(ln⁑2βˆ’ln⁑1)+(ln⁑(2+1)βˆ’ln⁑(1+1)) = -(\ln 2 - \ln 1) + (\ln (2 + 1) - \ln (1 + 1))

Simplifying further:

=βˆ’ln⁑2+0+ln⁑3βˆ’ln⁑2 = -\ln 2 + 0 + \ln 3 - \ln 2
  1. Final answer:
    Combine the terms:
=ln⁑3βˆ’ln⁑4=ln⁑34 = \ln 3 - \ln 4 = \ln \frac{3}{4}

Key Formulas or Methods Used

  • Substitution: u=1+cos⁑xu = 1 + \cos x, du=βˆ’sin⁑x,dxdu = -\sin x , dx
  • Logarithmic Integrals:
    • ∫1u,du=ln⁑∣u∣\int \frac{1}{u} , du = \ln |u|

Summary of Steps

  1. Make the substitution u=1+cos⁑xu = 1 + \cos x.
  2. Determine the new limits for uu.
  3. Substitute uu and dudu into the integral.
  4. Split the integral into two simpler parts.
  5. Integrate each part using logarithmic rules.
  6. Evaluate the integrals at the limits.
  7. Simplify the final expression to ln⁑34\ln \frac{3}{4}.