Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{0}^{\pi / 8} \cos 2 x d x$$

Short answer

Question: Evaluate the definite integral $$\int_{0}^{\pi / 8} \cos 2 x d x$$. Answer: The value of the definite integral is $$\frac{\sqrt{2}}{4}$$.

Step-by-step solution

Identify the function and its derivative within the integral

In this integral, the function is $$\cos(2x)$$. The derivative of $$2x$$ is $$2\,dx$$. The $$dx$$ is already present in our integral. For a substitution, let's consider the chain rule of derivatives, where the original function also includes a derivative inside it.

Make the substitution

Let's use the substitution method with $$u = 2x$$. We'll find the derivative of that, which is: $$\frac{d u}{d x} = 2$$ Now, we need to solve for $$d x$$, to replace it in the integral: $$d x = \frac{d u}{2}$$ Let's also find the new limits of integration. When $$x = 0$$: $$u = 2(0) = 0$$ And when $$x = \frac{\pi}{8}$$: $$u = 2\left(\frac{\pi}{8}\right) = \frac{\pi}{4}$$ Now, we can write the integral in terms of $$u$$: $$\int_{0}^{\pi / 4} \cos u \,\frac{d u}{2}$$

Evaluate the new integral

The new integral now consists of the basic integral of the cosine function. Let's get the anti-derivative of $$\cos u$$: $$\int \cos(u)\,d u = \sin(u) + C$$ Since we have a constant factor of $$\frac{1}{2}$$, it can be placed outside of the integral. So now we have: $$\frac{1}{2} \int_{0}^{\pi / 4} \cos u \,d u = \frac{1}{2}\left[\sin(u)\right]_{0}^{\pi/4}$$ Now let's evaluate for the limits: $$\frac{1}{2}\left[\sin\left(\frac{\pi}{4}\right) - \sin(0)\right]$$

Find the definite integral's value

Now, we just need to find the value of the sine function at the limits and subtract them: $$\frac{1}{2}\left[\frac{\sqrt{2}}{2} - 0\right]$$ Which simplifies to: $$\frac{\sqrt{2}}{4}$$ So the value of the definite integral is: $$\int_{0}^{\pi / 8} \cos 2 x d x = \frac{\sqrt{2}}{4}$$

Key concepts

Integration by Substitution

Understanding integration by substitution is crucial for solving a range of integral problems. It's very similar to the 'u-substitution' method learned in basic calculus. The idea is to simplify the integral into a form that we can easily integrate.

Here are the steps involved:

• Identify the inner function whose derivative also appears in the integral.
• Substitute this inner function with a new variable, typically u, and express dx in terms of du.
• Change the limits of integration to match the new variable.
• Perform the integration with respect to u.
• Finally, replace u with the original function to get the result in the original variable.

Performing these steps allows us to turn a complex integration problem into a simpler one that utilizes basic integration rules we're familiar with.

Cosine Function Integral

The cosine function, denoted as cos, is one of the fundamental trigonometric functions that we frequently encounter in integral calculus. The integral of the cosine function is straightforward: \
\[ \int \cos(x)\,dx = \sin(x) + C \]
where C represents the constant of integration. Whenever we encounter the cosine function within an integration problem, this fundamental antiderivative form helps simplify our calculations. As seen in our example exercise, the presence of a constant (2 in cos(2x)) can be handled through u-substitution, making the integral of cosine functions manageable irrespective of the constants involved.

Definite Integral Evaluation

Evaluating a definite integral involves calculating the area under the curve of a function within specified limits. First, we find the indefinite integral or the antiderivative. Then we apply the limits of integration, which involves substituting the upper limit into the antiderivative, subtracting the antiderivative evaluated at the lower limit. \
\[ F(b) - F(a) \]
This procedure gives us the net area between the function and the x-axis from a to b. In our example, the limits were transformed during the substitution step, which is a crucial stage in ensuring that we correctly evaluate the area corresponding to the original function's graph.