Key Concepts and Formulas

• Trigonometric Identity: $(|\sin x| + |\cos x|)^2 = \sin^2 x + \cos^2 x + 2|\sin x||\cos x| = 1 + |\sin 2x|$.
• Periodicity of Integrals: If a function $f(x)$ is periodic with period $T$, then $\int_0^{nT} f(x) \, dx = n \int_0^T f(x) \, dx$ for any integer $n$.
• Property of Absolute Value Integrals: $\int_a^b |f(x)| \, dx$ involves splitting the integral at the roots of $f(x)$ within the interval $[a, b]$.

Step-by-Step Solution

Step 1: Simplify the integrand. We are asked to evaluate the integral $\int_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x$. First, let's simplify the term $(|\sin x|+|\cos x|)^2$. Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$, we get: $(|\sin x|+|\cos x|)^2 = (|\sin x|)^2 + (|\cos x|)^2 + 2|\sin x||\cos x|$ Since $(|\sin x|)^2 = \sin^2 x$ and $(|\cos x|)^2 = \cos^2 x$, we have: $= \sin^2 x + \cos^2 x + 2|\sin x \cos x|$ Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and the double angle identity $\sin 2x = 2\sin x \cos x$, we get: $= 1 + |\sin 2x|$ So, the integral becomes $\int_{0}^{20 \pi}(1 + |\sin 2x|) d x$.

Step 2: Analyze the periodicity of the integrand. The integrand is $f(x) = 1 + |\sin 2x|$. The function $\sin 2x$ has a period of $\frac{2\pi}{2} = \pi$. The function $|\sin 2x|$ also has a period of $\pi$. To see this, $|\sin(2(x+\pi))| = |\sin(2x+2\pi)| = |\sin(2x)|$. Therefore, the function $1 + |\sin 2x|$ is periodic with a period $T = \pi$.

Step 3: Use the property of definite integrals over a period. The integral is from $0$ to $20\pi$. Since the period of the integrand is $\pi$, we are integrating over $n=20$ periods. We can use the property $\int_0^{nT} f(x) \, dx = n \int_0^T f(x) \, dx$. Here, $n=20$ and $T=\pi$. So, the integral can be rewritten as: $\int_{0}^{20 \pi}(1 + |\sin 2x|) d x = 20 \int_{0}^{\pi}(1 + |\sin 2x|) d x$.

Step 4: Evaluate the integral over one period. Now we need to evaluate $\int_{0}^{\pi}(1 + |\sin 2x|) d x$. This can be split into two integrals: $\int_{0}^{\pi} 1 \, dx + \int_{0}^{\pi} |\sin 2x| \, dx$.

The first part is straightforward: $\int_{0}^{\pi} 1 \, dx = [x]_0^\pi = \pi - 0 = \pi$.

For the second part, $\int_{0}^{\pi} |\sin 2x| \, dx$, we need to consider the sign of $\sin 2x$ in the interval $[0, \pi]$. Let $u = 2x$. When $x=0$, $u=0$. When $x=\pi$, $u=2\pi$. So, $du = 2 \, dx$, which means $dx = \frac{1}{2} du$. The integral becomes: $\int_{0}^{2\pi} |\sin u| \frac{1}{2} du = \frac{1}{2} \int_{0}^{2\pi} |\sin u| \, du$.

The function $|\sin u|$ is non-negative. In the interval $[0, 2\pi]$, $\sin u$ is positive for $u \in (0, \pi)$ and negative for $u \in (\pi, 2\pi)$. So, $\int_{0}^{2\pi} |\sin u| \, du = \int_{0}^{\pi} \sin u \, du + \int_{\pi}^{2\pi} (-\sin u) \, du$.

Let's evaluate these: $\int_{0}^{\pi} \sin u \, du = [-\cos u]_0^\pi = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$. $\int_{\pi}^{2\pi} (-\sin u) \, du = [\cos u]_\pi^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 1 + 1 = 2$.

Therefore, $\int_{0}^{2\pi} |\sin u| \, du = 2 + 2 = 4$. Substituting this back into the integral for $|\sin 2x|$: $\int_{0}^{\pi} |\sin 2x| \, dx = \frac{1}{2} \times 4 = 2$.

Now, combine the parts of the integral over one period: $\int_{0}^{\pi}(1 + |\sin 2x|) d x = \pi + 2$.

Step 5: Calculate the final integral value. We found that $\int_{0}^{\pi}(1 + |\sin 2x|) d x = \pi + 2$. The original integral is $20 \int_{0}^{\pi}(1 + |\sin 2x|) d x$. So, the value of the integral is $20 (\pi + 2)$.

Common Mistakes & Tips

• Forgetting the absolute value: If you forget the absolute value in $|\sin 2x|$, you might incorrectly assume $\sin 2x$ is always positive or integrate it directly, leading to an incorrect result.
• Incorrect period calculation: Ensure you correctly identify the period of $|\sin 2x|$. While $\sin 2x$ has period $\pi$, $|\sin 2x|$ also has period $\pi$.
• Integration limits for substitution: When performing a substitution like $u=2x$, be careful to change the integration limits accordingly.

Summary

The problem involves integrating a function containing absolute values of trigonometric terms over a large interval. We first simplified the integrand using trigonometric identities to get $1 + |\sin 2x|$. Recognizing that the integrand is periodic with period $\pi$, we used the property of definite integrals over a multiple of the period to reduce the problem to integrating over one period. We then evaluated the integral over one period by splitting it into parts and handling the absolute value by considering the sign of $\sin 2x$. Finally, we multiplied the result by the number of periods.

The final answer is $\boxed{20(\pi+2)}$.