Problem 93 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int_{0}^{\pi / 2} 7^{\cos t} \sin t d t\end{equation}

Step-by-Step Solution

Answer

\(\frac{6}{\ln 7}\)

Step 1: Identify the Integral Type

We have an integral of the form \(\int 7^{\cos t} \sin t \, dt\). The integral involves an exponential function raised to a trigonometric function, suggesting a substitution might be useful for evaluation.

Step 2: Choose an Appropriate Substitution

Choose the substitution \(u = \cos t\). Then, \(du = -\sin t \, dt\), or equivalently, \(\sin t \, dt = -du\). This substitution simplifies the integral in terms of \(u\). The limits of integration change accordingly: when \(t = 0\), \(u = \cos 0 = 1\); and when \(t = \pi/2\), \(u = \cos \pi/2 = 0\).

Step 3: Rewrite the Integral with New Limits

Substitute \(u = \cos t\) into the integral, which changes it to \(\int_{1}^{0} 7^u (-du)\). This simplifies to \(-\int_{1}^{0} 7^u \, du\), which can be re-expressed by switching the limits as \(\int_{0}^{1} 7^u \, du\).

Step 4: Evaluate the Simplified Integral

The integral \(\int_{0}^{1} 7^u \, du\) can be solved by recognizing it as an exponential integral. The integral of \(7^u\) with respect to \(u\) is \(\frac{7^u}{\ln 7}\). Evaluate this from \(u = 0\) to \(u = 1\) to get:\[\left[ \frac{7^u}{\ln 7} \right]_{0}^{1} = \frac{7^1}{\ln 7} - \frac{7^0}{\ln 7} = \frac{7}{\ln 7} - \frac{1}{\ln 7} = \frac{6}{\ln 7}.\]

Step 5: Write the Final Solution

The final value of the integral \(\int_{0}^{\pi/2} 7^{\cos t} \sin t \, dt\) is \(\frac{6}{\ln 7}\).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus that simplifies complex integrals. The main idea is to transform a challenging integral into a simpler one, by changing the variable of integration. This can be particularly helpful when dealing with integrals involving composite functions, such as exponential functions raised to trigonometric functions.In the exercise, we are working with the integral \( \int_{0}^{\pi / 2} 7^{\cos t} \sin t \, dt \). By recognizing that \( \cos t \) inside the exponential is related to the rest of the integrand, we choose \( u = \cos t \) for substitution. This choice is strategic because differentiating \( \cos t \) gives \( du = -\sin t \, dt \), which matches part of our integral.

• Identify a part of the function that, when differentiated, shows up elsewhere in the integrand.
• Make the substitution to transform the variable of integration from \( t \) to \( u \).
• Adjust the limits of integration to fit the new variable.

By using substitution, you transformed the original integral into \( \int_{0}^{1} 7^u \, du \) and made it significantly easier to solve.

Trigonometric Function

Trigonometric functions are fundamental in calculus, especially when computing integrals involving periodic behavior. In the context of the exercise, the function \(\cos t\) appears as the exponent of the base 7, making the function \(7^{\cos t}\).When dealing with these types of integrals:

• Recognize if a trigonometric function directly affects or modifies another function.
• Utilize trigonometric identities and relationships to simplify expressions when necessary.
• Focus on differentiating and integrating trigonometric functions accurately.

In our exercise, using \( u = \cos t \), we navigate through the trigonometric complexity by reducing the problem to a simpler exponential integral. This showcases how the understanding and manipulation of trigonometric functions are critical in solving integrals involving them.

Exponential Function

Dealing with exponential functions in integrals can initially seem daunting, but understanding their properties makes it much easier. An exponential function is expressed in the form \( a^u \), where \( a \) is a constant and \( u \) is the variable.To integrate exponential functions:

• Remember that \( \int a^u \, du = \frac{a^u}{\ln a} \), a key property used extensively in solving these integrals.
• Identify if the exponential is compounded with other functions, such as trigonometric ones.
• Apply substitution to simplify the integrand, as was done by setting \(u = \cos t\) in the exercise.

By converting the original trigonometric-exponential function \(7^{\cos t}\sin t\) into \(7^u\), we easily integrate to derive \(\frac{7^u}{\ln 7}\) evaluated from 0 to 1, resulting in the solution \(\frac{6}{\ln 7}\). This demonstrates the ease with which exponential functions can be resolved once simplified.