Question Statement

Evaluate the definite integral:

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}xdx$

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Background and Explanation

To integrate ${\cos }^{2}x$, we cannot use the power rule directly. Instead, we apply the power-reduction identity to rewrite ${\cos }^{2}x$ in terms of $\cos 2x$, which integrates easily using standard techniques.

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Solution

We begin by applying the trigonometric identity for ${\cos }^{2}x$:

$\begin{array}{rl}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}xdx& ={\int }_{-\pi /2}^{\pi /2}\frac{1+\cos 2x}{2}dx\left(\because {\cos }^{2}x=\frac{1+\cos 2x}{2}\right)\\ & =\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}(1+\cos 2x)dx\end{array}$

Next, we split the integral into two parts and integrate each term separately:

$\begin{array}{rl}& =\frac{1}{2}\left[{\int }_{-\pi /2}^{\pi /2}1dx+{\int }_{-\pi /2}^{\pi /2}\cos 2xdx\right]\\ & =\frac{1}{2}\left[{\left[x\right]}_{-\pi /2}^{\pi /2}+{\left[\frac{\sin 2x}{2}\right]}_{-\pi /2}^{\pi /2}\right]\end{array}$

Now we evaluate at the upper and lower limits. For the first term: $x$ evaluated from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$ gives $\frac{\pi }{2}-(-\frac{\pi }{2})=\pi$. For the second term, we substitute the bounds into $\frac{\sin 2x}{2}$:

$\begin{array}{rl}& =\frac{1}{2}\left[\left(\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right)+\frac{1}{2}\left(\sin (\pi )-\sin (-\pi )\right)\right]\\ & =\frac{1}{2}\left[\left(\frac{\pi }{2}+\frac{\pi }{2}\right)+\frac{1}{2}(0-0)\right](\because \sin (\pi )=0\text{ and }\sin (-\pi )=0)\\ & =\frac{1}{2}\left[\pi +0\right]\\ & =\frac{\pi }{2}\end{array}$

Therefore, the value of the integral is $\boxed{\frac{\pi }{2}}$.

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Key Formulas or Methods Used

• Power-reduction formula: ${\cos }^{2}x=\frac{1+\cos 2x}{2}$
• Constant integration: $\int 1dx=x+C$
• Cosine integration: $\int \cos (ax)dx=\frac{\sin (ax)}{a}+C$
• Sine values at multiples of $\pi$: $\sin (\pi )=\sin (-\pi )=0$
• Definite integral evaluation: ${\left[F(x)\right]}_a^b=F(b)-F(a)$

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Summary of Steps

1. Apply identity: Replace ${\cos }^{2}x$ with $\frac{1+\cos 2x}{2}$ and factor out $\frac{1}{2}$
2. Split integral: Separate into $\int 1dx$ and $\int \cos 2xdx$
3. Integrate: Find antiderivatives $x$ and $\frac{\sin 2x}{2}$ respectively
4. Evaluate bounds: Calculate ${\left[x\right]}_{-\pi /2}^{\pi /2}=\pi$ and $\frac{1}{2}[\sin 2x{]}_{-\pi /2}^{\pi /2}=0$
5. Combine: Multiply by the outer factor of $\frac{1}{2}$ to get $\frac{1}{2}(\pi +0)=\frac{\pi }{2}$