Problem 42
Question
Evaluate the definite integral. \(\int_{1 / 6}^{1 / 2} \csc \pi t \cot \pi t d t\)
Step-by-Step Solution
Verified Answer
The value of the definite integral is 1.
1Step 1: Set up the Integral
We need to evaluate the definite integral \( \int_{1/6}^{1/2} \csc(\pi t) \cot(\pi t) \, dt \). This integral involves the trigonometric functions \( \csc \) and \( \cot \), so it is helpful to recall that the derivative of \( \csc(x) \) is \(-\csc(x) \cot(x) \).
2Step 2: Identify an Antiderivative
Recognize that the integrand, \( \csc(\pi t) \cot(\pi t) \), is the derivative of \( -\csc(\pi t) \). So, an antiderivative of the integrand is \( -\csc(\pi t) \).
3Step 3: Evaluate the Antiderivative at the Bounds
Calculating the antiderivative at the upper bound, we get \( -\csc(\pi \times 1/2) = -\csc(\pi/2) \). Since \( \csc(\pi/2) = 1 \), this becomes \( -1 \).
4Step 4: Evaluate the Antiderivative at the Lower Bound
Calculate the antiderivative at the lower bound: \( -\csc(\pi \times 1/6) = -\csc(\pi/6) \). Since \( \csc(\pi/6) = 2 \), this becomes \( -2 \).
5Step 5: Calculate the Definite Integral
Substitute the values found in Steps 3 and 4 into the Fundamental Theorem of Calculus: \( -1 - (-2) \). Calculate this to get \( -1 + 2 = 1 \).
Key Concepts
Trigonometric FunctionsAntiderivativesFundamental Theorem of Calculus
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, especially in calculus. They describe relationships between the angles and sides of triangles. Two key trigonometric functions are used in this problem: the cosecant (\(\csc\)) and cotangent (\(\cot\)).
The function \(\csc(\pi t)\cot(\pi t)\) involves multiplying these two trigonometric expressions.
Because trigonometric functions can be transformed easily into their derivatives and antiderivatives, they are powerful tools in solving integrals.
- The cosecant of an angle is the reciprocal of its sine: \(\csc(x) = \frac{1}{\sin(x)}\).
- The cotangent of an angle is the reciprocal of its tangent: \(\cot(x) = \frac{1}{\tan(x)}\) or \(\cot(x) = \frac{\cos(x)}{\sin(x)}\).
The function \(\csc(\pi t)\cot(\pi t)\) involves multiplying these two trigonometric expressions.
Because trigonometric functions can be transformed easily into their derivatives and antiderivatives, they are powerful tools in solving integrals.
Antiderivatives
Antiderivatives are functions that reverse the process of differentiation. If you differentiate a function, an antiderivative would be the original function before differentiation.
In simpler terms, finding an antiderivative is just a way of determining what function could produce a given function when differentiated.
In this particular exercise, you recognize that the expression \(\csc(\pi t)\cot(\pi t)\) is the derivative of \(-\csc(\pi t)\).
By identifying this, it allows us to write \(-\csc(\pi t)\) as the antiderivative of the given function.
In simpler terms, finding an antiderivative is just a way of determining what function could produce a given function when differentiated.
In this particular exercise, you recognize that the expression \(\csc(\pi t)\cot(\pi t)\) is the derivative of \(-\csc(\pi t)\).
By identifying this, it allows us to write \(-\csc(\pi t)\) as the antiderivative of the given function.
- This step simplifies the integration process significantly because you can evaluate the antiderivative at the bounds.
- Being comfortable with identifying antiderivatives is crucial in solving integrals efficiently.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a critical concept that links differentiation and integration. It has two main components that simplify how we evaluate definite integrals:
In the problem at hand, once we identify the antiderivative \(-\csc(\pi t)\), we evaluated this expression at the upper bound (\(-\csc(\pi/2) = -1\)) and lower bound (\(-\csc(\pi/6) = -2\)).
By applying the Fundamental Theorem, you subtract the lower bound result from the upper bound result: \(-1 - (-2) = 1\).
This process confirms the power of the theorem in transforming a potentially complicated calculation into basic arithmetic.
- The first part asserts that if a function is continuous on a closed interval, then a function defined by an integral of that function is its antiderivative.
- The second part holds that the definite integral of a function over an interval is equal to the difference in the values of an antiderivative evaluated at the interval’s endpoints.
In the problem at hand, once we identify the antiderivative \(-\csc(\pi t)\), we evaluated this expression at the upper bound (\(-\csc(\pi/2) = -1\)) and lower bound (\(-\csc(\pi/6) = -2\)).
By applying the Fundamental Theorem, you subtract the lower bound result from the upper bound result: \(-1 - (-2) = 1\).
This process confirms the power of the theorem in transforming a potentially complicated calculation into basic arithmetic.
Other exercises in this chapter
Problem 41
Evaluate the definite integral. \(\int_{0}^{1} x^{2}\left(1+2 x^{3}\right)^{5} d x\)
View solution Problem 42
Find the general indefinite integral. \(\int \sec t(\sec t+\tan t) d t\)
View solution Problem 43
Find the general indefinite integral. \(\int \frac{\sin x}{1-\sin ^{2} x} d x\)
View solution Problem 43
Evaluate the definite integral. \(\int_{1}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x\)
View solution