Question Statement

Evaluate the definite integral:

${\int }_{\frac{\pi }{3}}^{\frac{\pi }{4}}\cos \theta d\theta$

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Background and Explanation

This problem requires knowledge of basic trigonometric integration and the Fundamental Theorem of Calculus. You need to recall that the antiderivative of $\cos \theta$ is $\sin \theta$, and how to evaluate definite integrals by substituting the upper and lower limits into the antiderivative.

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Solution

We evaluate the definite integral using the Fundamental Theorem of Calculus. The antiderivative of $\cos \theta$ is $\sin \theta$.

${\int }_{-\frac{\pi }{3}}^{\frac{\pi }{4}}\cos \theta d\theta ={\left[\sin \theta \right]}_{-\frac{\pi }{3}}^{\frac{\pi }{4}}$

Now we evaluate at the upper limit $\frac{\pi }{4}$ and subtract the evaluation at the lower limit $-\frac{\pi }{3}$:

$\begin{array}{rl}& =\sin \frac{\pi }{4}-\sin \left(\frac{-\pi }{3}\right)\\ & =\sin \frac{\pi }{4}+\sin \frac{\pi }{3}\\ & =\frac{1}{\sqrt{2}}+\left(\frac{\sqrt{3}}{2}\right)\\ & =\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\end{array}$

Thus, the value of the definite integral is $\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}$.

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Key Formulas or Methods Used

• $\int \cos \theta d\theta =\sin \theta +C$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is an antiderivative of $f$
• Odd function property of sine: $\sin (-\theta )=-\sin (\theta )$, which means $-\sin (-\frac{\pi }{3})=-(-\sin (\frac{\pi }{3}))=+\sin (\frac{\pi }{3})$
• Standard trigonometric values: $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$ and $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$

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Summary of Steps

1. Find the antiderivative of $\cos \theta$, which is $\sin \theta$
2. Apply the Fundamental Theorem of Calculus: evaluate ${\left[\sin \theta \right]}_{-\pi /3}^{\pi /4}=\sin (\frac{\pi }{4})-\sin (-\frac{\pi }{3})$
3. Apply the odd function property: $\sin (-\frac{\pi }{3})=-\frac{\sqrt{3}}{2}$, so subtracting yields addition: $\frac{1}{\sqrt{2}}-(-\frac{\sqrt{3}}{2})$
4. Substitute exact values to get the final result: $\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}$