91Ó°ÊÓ

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 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 \(\cot(x) = \frac{\cos(x)}{\sin(x)}\).

In the given exercise, both functions appear in the integrand.
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 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.

• 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.

Recognizing antiderivatives, therefore, builds the foundation to deal with definite integrals easily.

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:

• 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 simpler terms, it states:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]where \(F(x)\) is any antiderivative of \(f(x)\).
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.