Integrals of Sine and Cosine
Basic Properties
- Integral of Sine: The integral of the sine function, denoted as integral sin(x) dx, is equal to -cos(x) + C, where C is the constant of integration. Understanding how the differentiation of trigonometric functions works can provide deeper insights into the process of integration, specifically for sine and cosine functions. For more information, see Differentiation of Trigonometric Functions.
- Integral of Cosine: The integral of the cosine function, expressed as integral cos(x) dx, is equal to sin(x) + C.
These properties are foundational and will be utilised in various applications and examples throughout this section.
Example 1: Integrating Sine and Cosine Functions
Consider the integral integral sin(x) + cos(x) dx. To evaluate this, we can use the basic properties mentioned above:
integral sin(x) + cos(x) dx = -cos(x) + sin(x) + C
Example 2: Integrating with Limits
Evaluate integral from 0 to pi/2 sin(x) dx.
Using the fundamental theorem of calculus, we find the antiderivative of sin(x), which is -cos(x), and evaluate it from 0 to pi/2.
integral from 0 to pi/2 sin(x) dx = [-cos(x)] from 0 to pi/2 = [-cos(pi/2) - (-cos(0))] = [0 - (-1)] = 1
Example 3: Integrating Sine over a Half-Period
The integral of sine over a half-period, integral from 0 to pi sin(x) dx, is equal to 2. This can be visualised as the area under the curve of the sine function from 0 to pi, which is symmetric and equal to 2.
Example 4: Mean Square of Sine
The mean square of the sine function over a period, integral from 0 to 2 pi (sin2(x))/(2 pi) dx, is equal to 0.5. This represents the average value of the square of the sine function over one period. To enhance understanding of trigonometric identities, which play a significant role in the integration process, consider exploring Trigonometric Identities.
IB Maths Tutor Tip: Practising integration of trigonometric functions enhances your ability to solve complex problems, especially in physics, by understanding the relationship between angles, distances, and real-world phenomena.
Integral of sec2
Basic Property
- Integral of sec2: The integral of sec2(x), denoted as integral sec2(x) dx, is equal to tan(x) + C. his integral is pivotal in solving trigonometric equations, a fundamental skill in advanced mathematics. For further study, see Solving Trigonometric Equations.
Example 5: Integrating sec2
Consider the integral integral sec2(x) dx. Utilising the basic property:
integral sec2(x) dx = tan(x) + C
Example 6: Integrating sec2 with Limits
Evaluate integral from 0 to pi/4 sec2(x) dx.
We find the antiderivative of sec2(x), which is tan(x), and evaluate it from 0 to pi/4.
integral from 0 to pi/4 sec2(x) dx = [tan(x)] from 0 to pi/4 = [tan(pi/4) - tan(0)] = [1 - 0] = 1
Applications in Physics
Example 7: Calculating Work Done
In physics, the work done, W, by a force, F, over a distance, x, is given by the integral of the force function with respect to distance. Suppose a force is given by F(x) = sin(x) Newtons from x = 0 to x = pi metres. Find the work done.
W = integral from 0 to pi F(x) dx = integral from 0 to pi sin(x) dx = [-cos(x)] from 0 to pi = [-cos(pi) - (-cos(0))] = [1 - (-1)] = 2 Joules
IB Tutor Advice: For exams, ensure you're comfortable with the fundamental integrals of trigonometric functions and their applications in physics to confidently tackle both theoretical questions and practical problems.
Example 8: Evaluating Displacement
In kinematics, displacement, s, can be determined by integrating the velocity function, v(t), with respect to time, t. If v(t) = cos(t) m/s from t = 0 to t = pi seconds, find the displacement.
s = integral from 0 to pi v(t) dt = integral from 0 to pi cos(t) dt = [sin(t)] from 0 to pi = [sin(pi) - sin(0)] = [0 - 0] = 0 metres. This example illustrates the application of integration in physics, specifically in calculating displacement using trigonometric functions. For an in-depth exploration of integration techniques, including those applied to exponential and logarithmic functions, consider visiting Integration of Exponential and Logarithmic Functions.
In this section, we have meticulously explored the integrals of sine, cosine, and sec2, providing a foundational understanding of how to evaluate these integrals and apply them in practical contexts, such as physics. Through detailed examples, we have demonstrated the application of basic integration properties and rules, ensuring a comprehensive understanding of the topic. For further refinement of your integration skills, particularly in varied contexts, Techniques of Integration offers valuable insights. Remember to always verify your results and practice with various problems to solidify your knowledge and skills in integration.
FAQ
The integral of sin2(x) typically cannot be found without using trigonometric identities due to the square on the sine function, which complicates the integration. The standard method involves using the double-angle identity: sin2(x) = (1 - cos(2x))/2. This identity allows us to express sin2(x) in a form that can be easily integrated term by term. Without using this identity, we do not have a straightforward antiderivative for sin2(x) in the standard integral tables, and direct integration becomes non-trivial. The use of trigonometric identities, in this case, simplifies the integral into a form that can be readily evaluated.
The integral of sec(x) involves the natural logarithm due to the method of integration used to evaluate it, which is integration by substitution. The integral of sec(x) dx can be evaluated by multiplying and dividing the integrand by (sec(x) + tan(x)) and then letting u = sec(x) + tan(x). So, integral sec(x) dx = integral sec(x) * (sec(x) + tan(x))/(sec(x) + tan(x)) dx. With the substitution u = sec(x) + tan(x), du = (sec(x)tan(x) + sec2(x)) dx, the integral becomes integral du/u, which is ln|u| + C. Substituting back, we get ln|sec(x) + tan(x)| + C. The natural logarithm arises due to the form of the integral after substitution, which is a standard integral form.
The integral of tan(x) can be determined by recognising that tan(x) = sin(x)/cos(x). So, integral tan(x) dx = integral sin(x)/cos(x) dx. A common technique involves substituting u = cos(x), and thus du = -sin(x) dx. The integral becomes -integral du/u, which is -ln|u| + C. Substituting back, we get -ln|cos(x)| + C as the antiderivative of tan(x). Geometrically, the integral from a to b of tan(x) dx represents the net area between the curve y = tan(x) and the x-axis from x = a to x = b. This could be interpreted as the total "displacement" of the tan(x) function from the x-axis over that interval.
The period of a trigonometric function significantly impacts its integral by influencing the net area under the curve over one or more periods. For functions like sin(x) and cos(x), which have a period of 2π, the integral over one full period (from 0 to 2π, for instance) is zero because the positive and negative areas cancel each other out. However, if we consider a half-period or any interval that doesn’t complete a full cycle, the integral is non-zero and represents the net area (or “net displacement”) under the curve over that interval. Understanding the period is crucial for interpreting integrals of trigonometric functions, especially in applications like physics, where the integral might represent physical quantities like displacement or work done.
The integrals of cos2(x) and sin2(x) might seem different but they are evaluated using a similar trigonometric identity. The double-angle identities are used here: cos2(x) = (1 + cos(2x))/2 and sin2(x) = (1 - cos(2x))/2. When you integrate each from a to b, you'll apply these identities first and then integrate term by term. For cos2(x), integral cos2(x) dx = integral (1 + cos(2x))/2 dx = (x/2) + (sin(2x)/4) + C. For sin2(x), integral sin2(x) dx = integral (1 - cos(2x))/2 dx = (x/2) - (sin(2x)/4) + C. Both integrals involve a linear term and a sinusoidal term, providing a blend of algebraic and trigonometric elements in their antiderivatives.
Practice Questions
The integral from 0 to pi/2 of sin(x) dx can be evaluated by finding the antiderivative of sin(x), which is -cos(x), and then applying the Fundamental Theorem of Calculus. So, integral from 0 to pi/2 sin(x) dx = [-cos(x)] from 0 to pi/2 = [-cos(pi/2) - (-cos(0))] = [0 - (-1)] = 1. The result, 1, represents the area under the curve of the sine function from 0 to pi/2. This can be visualised as the space enclosed between the sine wave and the x-axis over this interval, which is a measure of the total positive excursion of the sine function from 0 to pi/2.
To find the total displacement of the particle from t = 0 to t = pi seconds, we need to integrate the velocity function, v(t) = cos(t), with respect to time over the given interval. So, s = integral from 0 to pi v(t) dt = integral from 0 to pi cos(t) dt. Finding the antiderivative of cos(t), which is sin(t), and applying the limits, we get s = [sin(t)] from 0 to pi = [sin(pi) - sin(0)] = [0 - 0] = 0 metres. Thus, the total displacement of the particle from t = 0 to t = pi seconds is 0 metres, meaning the particle returns to its starting position after pi seconds.
