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Notely Maths 12
Class 12 Maths
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3.6 Q-16

Question Statement

Evaluate the integral: ∫0π6cos⁡3θ,dθ\int_{0}^{\frac{\pi}{6}} \cos^3 \theta , d\theta

Background and Explanation

This problem involves integrating a trigonometric function, specifically cos⁡3θ\cos^3 \theta. To simplify the integration, we can use the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta, which will allow us to express the integrand in a form that is easier to work with. Understanding basic trigonometric identities and integration techniques, such as the integration of powers of sine and cosine, is necessary here.

Solution

  1. Rewrite the Expression:

    Start by breaking cos⁡3θ\cos^3 \theta into cos⁡2θ⋅cos⁡θ\cos^2 \theta \cdot \cos \theta: ∫0π6cos⁡3θ,dθ=∫0π6cos⁡2θ⋅cos⁡θ,dθ\int_{0}^{\frac{\pi}{6}} \cos^3 \theta , d\theta = \int_{0}^{\frac{\pi}{6}} \cos^2 \theta \cdot \cos \theta , d\theta

    Use the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to simplify: =∫0π6(1−sin⁡2θ)cos⁡θ,dθ= \int_{0}^{\frac{\pi}{6}} (1 - \sin^2 \theta) \cos \theta , d\theta

  2. Split the Integral:

    Distribute cos⁡θ\cos \theta over the terms inside the parentheses: =∫0π6cos⁡θ,dθ−∫0π6sin⁡2θ⋅cos⁡θ,dθ= \int_{0}^{\frac{\pi}{6}} \cos \theta , d\theta - \int_{0}^{\frac{\pi}{6}} \sin^2 \theta \cdot \cos \theta , d\theta

  3. Integrate the First Term:

    The integral of cos⁡θ\cos \theta is straightforward: ∫cos⁡θ,dθ=sin⁡θ\int \cos \theta , d\theta = \sin \theta

    So, evaluating from 00 to π6\frac{\pi}{6}: sin⁡(π6)−sin⁡(0)=12−0=12\sin \left(\frac{\pi}{6}\right) - \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}

  4. Integrate the Second Term:

    For sin⁡2θ⋅cos⁡θ\sin^2 \theta \cdot \cos \theta, use substitution: Let u=sin⁡θu = \sin \theta, so du=cos⁡θ,dθdu = \cos \theta , d\theta. The limits change accordingly:

    • When θ=0\theta = 0, u=0u = 0
    • When θ=π6\theta = \frac{\pi}{6}, u=sin⁡(π6)=12u = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}

    The integral becomes: ∫0π6sin⁡2θ⋅cos⁡θ,dθ=∫012u2,du\int_{0}^{\frac{\pi}{6}} \sin^2 \theta \cdot \cos \theta , d\theta = \int_{0}^{\frac{1}{2}} u^2 , du

    The integral of u2u^2 is: ∫u2,du=u33\int u^2 , du = \frac{u^3}{3}

    Evaluating from 00 to 12\frac{1}{2}: 13[(12)3−03]=13×18=124\frac{1}{3} \left[\left(\frac{1}{2}\right)^3 - 0^3\right] = \frac{1}{3} \times \frac{1}{8} = \frac{1}{24}

  5. Combine the Results:

    Now, combine the two terms: 12−124\frac{1}{2} - \frac{1}{24}

    Simplifying: 1224−124=1124\frac{12}{24} - \frac{1}{24} = \frac{11}{24}

Key Formulas or Methods Used

  • Trigonometric Identity: cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta

  • Basic Integration: ∫cos⁡θ,dθ=sin⁡θ\int \cos \theta , d\theta = \sin \theta

  • Substitution Method: When dealing with sin⁡2θ⋅cos⁡θ\sin^2 \theta \cdot \cos \theta, use substitution u=sin⁡θu = \sin \theta to simplify the integral.

Summary of Steps

  1. Use the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to simplify the integral.
  2. Split the integral into two parts.
  3. Integrate cos⁡θ\cos \theta directly.
  4. Use substitution to integrate sin⁡2θ⋅cos⁡θ\sin^2 \theta \cdot \cos \theta.
  5. Combine the results to get the final answer: 1124\frac{11}{24}