3.6 Q-11

Question Statement

Evaluate the integral: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos(t) , dt$

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Background and Explanation

To solve this integral, we need to recognize that the integral of $\cos(t)$ is a standard antiderivative. Specifically: $\int \cos(t) , dt = \sin(t) + C$

Thus, we will integrate $\cos(t)$ and then apply the limits of integration.

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Solution

1. Recognize the Standard Integral:

   The given integral is: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos(t) , dt$

   Using the standard formula for the integral of $\cos(t)$: $\int \cos(t) , dt = \sin(t)$

2. Apply the Limits of Integration:

   Now, we evaluate $\sin(t)$ at the limits $\frac{\pi}{3}$ and $\frac{\pi}{6}$: $\sin(t) \bigg|_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

   This means we need to compute: $\sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right)$

3. Compute the Sine Values:

   From trigonometric values, we know: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

   Substituting these values, we get: $\frac{\sqrt{3}}{2} - \frac{1}{2}$

4. Simplify the Result:

   Simplifying the expression: $\frac{\sqrt{3} - 1}{2}$

   Therefore, the value of the integral is: $\boxed{\frac{\sqrt{3} - 1}{2}}$

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Key Formulas or Methods Used

• Integral of Cosine:
  $\int \cos(t) , dt = \sin(t)$

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Summary of Steps

1. Recognize that $\int \cos(t) , dt = \sin(t)$.

2. Apply the limits of integration: $\sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right)$

3. Use known trigonometric values: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

4. Simplify the result to: $\frac{\sqrt{3} - 1}{2}$

5. Final answer: $\boxed{\frac{\sqrt{3} - 1}{2}}$