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3.6 Q-29

Question Statement

Evaluate the integral:

βˆ«Ο€/6Ο€/2Cos⁑xSin⁑x(2+Sin⁑x),dx\int_{\pi / 6}^{\pi / 2} \frac{\operatorname{Cos} x}{\operatorname{Sin} x (2 + \operatorname{Sin} x)} , dx

Background and Explanation

To solve this integral, we need to understand the basic trigonometric identities and properties. Specifically:

  • Cosecant (Cosec⁑x=1sin⁑x\operatorname{Cosec} x = \frac{1}{\sin x}) and cotangent (Cot⁑x=cos⁑xsin⁑x\operatorname{Cot} x = \frac{\cos x}{\sin x}) will help simplify the expression.
  • The integral involves a rational trigonometric expression, which requires manipulating and simplifying using these identities.
  • This type of integral can often be simplified by substitution or recognizing standard integral forms.

Solution

Step-by-step explanation:

  1. Rewrite the integrand:
    We start by rewriting the integrand using trigonometric identities.
Cos⁑xSin⁑x(2+Sin⁑x)=(2sin⁑x+1)βˆ’1(βˆ’2Cosec⁑xCot⁑x) \frac{\operatorname{Cos} x}{\operatorname{Sin} x (2 + \operatorname{Sin} x)} = \left( \frac{2}{\sin x} + 1 \right)^{-1} (-2 \operatorname{Cosec} x \operatorname{Cot} x)
  1. Substitute:
    Next, we simplify the expression by noting that Cosec⁑x=1sin⁑x\operatorname{Cosec} x = \frac{1}{\sin x} and Cot⁑x=cos⁑xsin⁑x\operatorname{Cot} x = \frac{\cos x}{\sin x}. This leads to:
βˆ«Ο€/6Ο€/2(2Cosec⁑x+1)βˆ’1(βˆ’2Cosec⁑xCot⁑x),dx \int_{\pi / 6}^{\pi / 2} (2 \operatorname{Cosec} x + 1)^{-1} (-2 \operatorname{Cosec} x \operatorname{Cot} x) , dx
  1. Factor and simplify:
    We can simplify the expression further, and focus on the integration:
=βˆ’16βˆ«Ο€/6Ο€/2(2Cosec⁑x+1)βˆ’1(βˆ’2Cosec⁑xCot⁑x),dx = -\frac{1}{6} \int_{\pi / 6}^{\pi / 2} (2 \operatorname{Cosec} x + 1)^{-1} (-2 \operatorname{Cosec} x \operatorname{Cot} x) , dx
  1. Apply integration rules:
    Using standard integral rules for logarithmic and trigonometric functions, we get:
=βˆ’12ln⁑((2Cosec⁑x+1))βˆ£Ο€/6Ο€/2 = -\frac{1}{2} \ln \left( (2 \operatorname{Cosec} x + 1) \right) \bigg|_{\pi / 6}^{\pi / 2}
  1. Evaluate the bounds:
    Finally, evaluate the logarithmic function at the limits x=Ο€/2x = \pi / 2 and x=Ο€/6x = \pi / 6:
=βˆ’12[ln⁑(2sin⁑(Ο€/2)+1)βˆ’ln⁑(2sin⁑(Ο€/6)+1)] = -\frac{1}{2} \left[ \ln \left( \frac{2}{\sin (\pi / 2)} + 1 \right) - \ln \left( \frac{2}{\sin (\pi / 6)} + 1 \right) \right]

Simplifying further:

=βˆ’12[ln⁑(2+1)βˆ’ln⁑(4+1)] = -\frac{1}{2} \left[ \ln (2 + 1) - \ln (4 + 1) \right]
  1. Final answer:
    This results in:
=12[ln⁑3βˆ’ln⁑5]=12ln⁑53 = \frac{1}{2} \left[ \ln 3 - \ln 5 \right] = \frac{1}{2} \ln \frac{5}{3}

Key Formulas or Methods Used

  • Trigonometric Identities:
    • Cosec⁑x=1sin⁑x\operatorname{Cosec} x = \frac{1}{\sin x}
    • Cot⁑x=cos⁑xsin⁑x\operatorname{Cot} x = \frac{\cos x}{\sin x}
  • Integral of Logarithms:
    • The integral of the form ∫dxx=ln⁑∣x∣\int \frac{dx}{x} = \ln |x|

Summary of Steps

  1. Rewrite the integrand using trigonometric identities.
  2. Simplify the integrand and apply the integral formula for logarithms.
  3. Evaluate the integral at the limits x=Ο€/6x = \pi / 6 and x=Ο€/2x = \pi / 2.
  4. Simplify the result to obtain 12ln⁑53\frac{1}{2} \ln \frac{5}{3}.