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Magazine Chapter 7: Problem 39
Evaluate the integral. $$ \int 3^{\cos x} \sin x d x $$
Short Answer
Expert verified
\(-\frac{3^{\cos x}}{\ln 3} + C\)
Step by step solution
01
Recognize the Structure
The integral \( \int 3^{\cos x} \sin x \, dx \) features a composition of functions: an exponential function and a trigonometric function. We will use substitution to simplify this integral.
02
Choose Appropriate Substitution
Let \( u = \cos x \), which makes \( \frac{du}{dx} = -\sin x \). Thus, \( du = -\sin x \, dx \). This allows us to rewrite the integral in terms of \( u \).
03
Substitute and Simplify
The integral becomes \( \int 3^u (-du) \), which simplifies to \( - \int 3^u \, du \). This adjustment makes the integration process more straightforward.
04
Integrate with Respect to \( u \)
The integral \(- \int 3^u \, du\) can be evaluated as follows: the antiderivative of \( a^u \) is \( \frac{a^u}{\ln a} \), provided \( a > 0 \). Applying this to our problem, \(- \int 3^u \, du = - \frac{3^u}{\ln 3} + C\).
05
Substitute Back to Original Variable
Re-substitute \( u = \cos x \) back into the expression: \(- \frac{3^{\cos x}}{\ln 3} + C\). This expression represents the original indefinite integral in terms of \( x \).
06
Conclude the Solution
Having performed all substitutions and integrations, the final solution of the integral is: \(- \frac{3^{\cos x}}{\ln 3} + C\), where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used in integral calculus to simplify the integration of complex expressions. It involves replacing a part of the integral with a new variable to reduce the integrand to a more manageable form.
In the integral \( \int 3^{\cos x} \sin x \, dx \), we face a compound function that involves both an exponential and a trigonometric function. By using substitution, we aim to break down this complexity.
In the integral \( \int 3^{\cos x} \sin x \, dx \), we face a compound function that involves both an exponential and a trigonometric function. By using substitution, we aim to break down this complexity.
- Identify a substitution: For this problem, we select \( u = \cos x \) because it simplifies the interaction between \( \cos x \) and its derivative, \( \sin x \).
- Calculate the differential: Once the substitution is made, find the derivative of \( u \) with respect to \( x \), giving \( \frac{du}{dx} = -\sin x \), or equivalently \( du = -\sin x \, dx \).
- Rewrite the integral: Substitute \( u \) and \( du \) into the integral, transforming it from a complex trigonometric-exponential form into \( -\int 3^u \, du \).
Trigonometric Functions
Trigonometric functions are fundamental in calculus. They include \( \sin x \), \( \cos x \), and \( \tan x \), among others. These functions often involve periodic behavior, making them crucial for understanding wave patterns and circular motion.
- In our integral, \( \cos x \) is part of the exponent, showing how trigonometric functions can appear in various mathematical contexts beyond simple wave functions.
- \( \sin x \) plays the role of the derivative here, and importantly, accompanies the function we are integrating. This relationship between \( \cos x \) and \( \sin x \) is critical in simplifying through substitution, especially as they are derivatives of each other, allowing proper alignment during substitution.
Exponential Functions
Exponential functions are characterized by a constant raised to a variable power. They are crucial in many areas, such as growth models and decay processes. In the context of the exercise, the function \( 3^{\cos x} \) represents an exponential function where the exponent is a trigonometric expression, adding complexity.
When integrating exponential functions like \( 3^u \), it鈥檚 key to apply the formula for the antiderivative of an exponential function: \( \frac{a^u}{\ln a} \), where \( a \) is a positive constant.
When integrating exponential functions like \( 3^u \), it鈥檚 key to apply the formula for the antiderivative of an exponential function: \( \frac{a^u}{\ln a} \), where \( a \) is a positive constant.
- This formula helps us efficiently handle integrations of the form \( \int a^u \, du \).
- In our problem, substituting \( u = \cos x \) made \( 3^{\cos x} \sin x \, dx \) more manageable as \( 3^u \, du \), utilizing the simple antiderivative of exponential functions.
Antiderivatives
Antiderivatives, also known as indefinite integrals, represent the inverse operation of taking a derivative. They provide a family of functions whose derivative is the integrand. Identifying the antiderivative is a vital step in solving integrals.
For exponential functions like \( a^u \), the antiderivative is \( \frac{a^u}{\ln a} \). In this particular problem, we determine the antiderivative of \( 3^u \) to be \( \frac{3^u}{\ln 3} \).
For exponential functions like \( a^u \), the antiderivative is \( \frac{a^u}{\ln a} \). In this particular problem, we determine the antiderivative of \( 3^u \) to be \( \frac{3^u}{\ln 3} \).
- Once you've found the antiderivative with respect to the substituted variable \( u \), don't forget to return to the original variable \( x \). This is achieved by resubstituting \( u = \cos x \) back into the expression.
- The final step includes adding a constant \( C \), which is important in all indefinite integrals as they denote the general solution that fits a family of specific solutions.
