Question

$$\text {Evaluate the following integrals.}$$ $$\int_{-\pi / 4}^{\pi / 4} \sqrt{1+\cos 4 x} d x$$

Short answer

Question: Find the value of the following integral: $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sqrt{1+\cos(4x)}}{\sin(4x)}dx$$ Answer: The value of the given integral is 0.

Step-by-step solution

Apply substitution

Let's substitute $$u = \cos 4x$$ so, $$\frac{du}{dx} = -4\sin(4x)$$ which implies $$dx = -\frac{1}{4} \frac{du}{\sin(4x)}$$. The integral becomes: $$-\frac{1}{4}\int \sqrt{1+u} \, \frac{du}{\sin(4x)}$$.

Find bounds for u

We also need to find the bounds for u. When $$x=-\frac{\pi}{4}$$: $$u = \cos\left(4\left(-\frac{\pi}{4}\right)\right) = \cos(-\pi) = -1$$. When $$x=\frac{\pi}{4}$$: $$u = \cos\left(4\left(\frac{\pi}{4}\right)\right) = \cos(\pi) = -1$$. So the bounds for u are -1 to -1.

Evaluate the integral

Since the bounds for u are the same (-1 to -1), the integral is 0: $$-\frac{1}{4}\int_{-1}^{-1} \sqrt{1+u} \, \frac{du}{\sin(4x)} = 0$$. Therefore, the value of the given integral is 0.

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus, used to determine the area under a curve or to accumulate quantities. In this exercise, we employ integration techniques to solve a definite integral. These techniques often involve a substitution method, which simplifies the integral by changing variables.

When approaching an integral, identifying a suitable substitution is key. In our example, we replaced the trigonometric expression using a substitution, transforming the original integral into a new form that can be more manageable.

Basic integration techniques include:

• Substitution: Useful for integrals involving composite functions. It simplifies the expression by changing the variable.
• Integration by Parts: Used when our integrand is a product of functions, defined by the rule \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Helps integrate rational functions by expressing the integrand as a sum of simpler fractions.

In our problem, substitution is particularly helpful because it deals with trigonometric functions, easing the calculation of the definite integral.

Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals involving square roots and trigonometric functions. This method is particularly effective when dealing with expressions like \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\).

In the given exercise, we encountered \(\sqrt{1+\cos 4x}\). Recognizing this as a candidate for trigonometric substitution, we used the identity for cosine to simplify the integral's complexity.

The use of trigonometric substitution generally involves:

• Choosing an appropriate substitution that converts the trigonometric function into algebraic form.
• Adjusting the integral's bounds if it's a definite integral, as seen in our solution.
• Handling the differential part of the substitution correctly, to ensure the new integral is in terms of the selected variable.

In our solution, the change \(u = \cos 4x\) simplified the integral, leading us to explore the bounds and transform the problem into a more straightforward evaluation.

Bounds of Integration

The bounds of integration are integral to evaluating a definite integral. They tell us the interval over which to integrate, often representing endpoints or limits of a function.

In definite integrals, once a substitution is made, the bounds also transform accordingly. This step is crucial to ensuring the new integral is properly set up for evaluation.

In our exercise, the bounds were initially given as \(-\pi/4\) to \(\pi/4\). Upon utilizing trigonometric substitution, we also needed to update these bounds to match our new variable \(u\).

Key points about bounds include:

• Always calculate the bounds based on the original conditions of the problem.
• Update bounds accurately when variables are substituted, ensuring they relate to the new form of the integral.
• Special cases: If the bounds are identical, as seen here where both became -1, the integral equals zero because the area between identical limits is zero.

Correctly working with bounds after substitution helped us realize the integral in the exercise evaluates to zero due to its same starting and ending points.