Question

\begin{aligned} &\int_{0}^{\pi / 2} \frac{\cos 2 x}{\cos x+\sin x} d x \text { equals }\\\ &\begin{array}{llll} (\text { a })-1 & \text { (b) } 0 & \text { (c) } 1 & \text { (d) } 2 \end{array} \end{aligned}

Short answer

The integral evaluates to 0.

Step-by-step solution

Simplify the Integrand with Trigonometric Identity

The integrand is \( \frac{\cos 2x}{\cos x + \sin x} \). Recall the trigonometric identity for \( \cos 2x = \cos^2 x - \sin^2 x \). This can be factored using \( \cos 2x = (\cos x + \sin x)(\cos x - \sin x) \). Substituting this into the integrand, we have: \[ \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x. \] Thus, the integral simplifies to \( \int_0^{\pi/2} (\cos x - \sin x) \, dx \).

Integrate Simplified Function

Now, integrate the simplified function \( \int_{0}^{\pi/2} (\cos x - \sin x) \, dx \). The integral of \( \cos x \) is \( \sin x \), and the integral of \( -\sin x \) is \( \cos x \). Therefore, the integral becomes: \[ \int_{0}^{\pi/2}(\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_0^{\pi/2}. \]

Evaluate Definite Integral at Boundaries

Evaluate the result of the integration bound from 0 to \( \pi/2 \). This gives: \[ \left[ \sin x + \cos x \right]_0^{\pi/2} = (\sin(\pi/2) + \cos(\pi/2)) - (\sin(0) + \cos(0)). \] Calculate each term: \( \sin(\pi/2) = 1 \), \( \cos(\pi/2) = 0 \), \( \sin(0) = 0 \), and \( \cos(0) = 1 \). So the expression becomes \( (1 + 0) - (0 + 1) = 0 \).

Key concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that relate different trigonometric functions to one another. When integrating problems that involve trigonometric functions, these identities often help to simplify the problem into a more manageable form.
For instance, in the given exercise, the identity \( \cos 2x = \cos^2 x - \sin^2 x \) was used to rewrite the integrand. This specific identity is one of the double-angle formulas. These are incredibly helpful because they allow us to express trigonometric functions at doubled angles as functions of the original angle. In our step-by-step solution, this identity was further factored to \( \cos 2x = (\cos x + \sin x)(\cos x - \sin x) \).
Using identities like this one can often turn complex rational expressions, which might be unwieldy to integrate directly, into simpler forms that are much more straightforward to handle, such as sums or differences of basic trigonometric functions.

Definite Integrals

Definite integrals are used to calculate the net area under a curve, from a specific starting point to an ending point. These integrals are bounded by limits, known as the lower limit and upper limit of integration. In this problem, the definite integral is evaluated from \( 0 \) to \( \pi/2 \).
Once you have set up your integral with its limits, the fundamental theorem of calculus helps out. It states that if you find the antiderivative of the function and evaluate it at these limits, you will get your answer. So after applying the trigonometric identity and simplifying the integrand, the function becomes \( \int_0^{\pi/2} (\cos x - \sin x) \, dx \).
The process involves finding the antiderivatives: \( \sin x \) is the antiderivative of \( \cos x \), and \( \cos x \) is the antiderivative of \(-\sin x\). Then, substitute the upper and lower limits into these antiderivatives and take the difference to find the answer.

Cosine and Sine Functions

The cosine and sine functions are fundamental in trigonometry. They describe the shape of a wave which cycles every \( 2\pi \) radians. In mathematical terms, they are periodic functions that are central to oscillatory behavior.
In the integration problem at hand, both of these functions appear after the integrand has been simplified. By knowing that the integral of \( \cos x \) is \( \sin x \) and the integral of \( -\sin x \) is \( \cos x \), we can find the antiderivatives during integration.
Another interesting property is how these functions behave at key angles. For instance, at \( x = 0 \) and \( x = \pi/2\), \( \sin \) and \( \cos \) have specific values: at \( \pi/2 \), \( \sin \) is at its maximum of 1 and \( \cos \) is 0. Conversely, at 0, \( \sin \) is 0 and \( \cos \) is 1. These values are crucial when evaluating definite integrals, as they provide the yield for our integral's boundaries.