Question

Evaluate the definite integral. $$ \int_{-\pi / 2}^{\pi / 2} \cos ^{3} x d x $$

Short answer

The definite integral \( \int_{-\pi / 2}^{\pi / 2} \cos ^{3} x \, dx \) is equal to \( \frac{1}{2} \).

Step-by-step solution

Transform the integral

The integral of \( \cos ^{3} x \) is a bit difficult to handle directly. This exercise can be simplified by using the following power-reduction identity: \[ \cos^{3}x = \frac{1}{4}\cos(x) + \frac{3}{4}\cos(3x) \]. Substituting into the original integral gives us \[ \int_{-\pi / 2}^{\pi / 2} \cos ^{3} x \, dx = \int_{-\pi / 2}^{\pi / 2} \left(\frac{1}{4}\cos(x) + \frac{3}{4}\cos(3x)\right) \, dx \].

Break down the integral into simpler parts

The linearity of the integral allows us to break down the integral into simpler parts which can be handled individually: \[ \int_{-\pi / 2}^{\pi / 2} \left(\frac{1}{4}\cos(x) + \frac{3}{4}\cos(3x)\right) dx = \frac{1}{4}\int_{-\pi / 2}^{\pi / 2} \cos(x) dx + \frac{3}{4}\int_{-\pi / 2}^{\pi / 2} \cos(3x) dx \]

Calculate the integrals

Now, each integral can be calculated individually. The integral of \(\cos(x)\) is \(\sin(x)\), and the integral of \(\cos(3x)\) is \( \frac{1}{3}\sin(3x) \): \[ \frac{1}{4}\int_{-\pi / 2}^{\pi / 2} \cos(x) dx = \frac{1}{4} [\sin(x)]_{-\pi / 2}^{\pi / 2} = \frac{1}{2} \] \[ \frac{3}{4}\int_{-\pi / 2}^{\pi / 2} \cos(3x) dx = \frac{3}{4} [\frac{1}{3}\sin(3x)]_{-\pi / 2}^{\pi / 2} = 0 \]

Combine the results

Finally, we combine the results obtained in step 3. The Integral is equal to the sum of these results: \[ \frac{1}{2} + 0 = \frac{1}{2} \].