/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Find the general antiderivative.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general antiderivative. Check your answers by differentiation. $$f(x)=e^{\sin x} \cos x$$

Short Answer

Expert verified
The general antiderivative is \(F(x) = e^{\sin x} + C\).

Step by step solution

01

Identify the Integral

The given function is \(f(x) = e^{\sin x} \cos x\). We need to find its antiderivative. Notice that this function appears suited for a substitution method.
02

Choose a Substitution

To simplify the integration process, let's substitute \(u = \sin x\). This implies that \(du = \cos x \, dx\). Thus, the integra becomes \(\int e^{u} \, du\).
03

Integrate Using Substitution

The integral becomes \(\int e^{u} \, du\), which is a simple integral with respect to \(u\). The antiderivative of \(e^u\) is \(e^u + C\).
04

Substitute Back

Now we substitute back the original variable: since \(u = \sin x\), we have the antiderivative of the original function as \(F(x) = e^{\sin x} + C\).
05

Verify by Differentiation

To check our work, differentiate \(F(x) = e^{\sin x} + C\). Using the chain rule, the derivative is \(\frac{d}{dx}[e^{\sin x} + C] = e^{\sin x} \cdot \cos x = f(x)\). This confirms that our solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
An antiderivative is a function that "undoes" the process of differentiation. If you have a function \(f(x)\), finding its antiderivative means determining a function \(F(x)\) such that \(F'(x) = f(x)\). The process of finding the antiderivative is also known as integration.When you integrate a function and find its antiderivative, you usually add a constant of integration ("+ C") at the end. This constant accounts for the fact that there is an infinite number of antiderivatives for any given function, differing only by a constant.In our example with \(f(x) = e^{\sin x} \cos x\), the antiderivative is \(F(x) = e^{\sin x} + C\). We can confirm this by differentiating \(F(x)\) and checking if it equals \(f(x)\). If it does, we've found the correct antiderivative.
Substitution Method
The substitution method is a handy technique used to simplify the process of integration. It's especially useful for integrals involving composite functions. The idea is to choose a substitution that will transform the integral into a simpler form that is easier to solve.Here’s how substitution works:
  • You pick a part of the integrand (the expression being integrated) to substitute with a new variable.
  • You express the differential ("\(dx\)") and the integral in terms of this new variable.
  • Integrate with respect to this new variable.
  • Finally, substitute back the original variable to get your final answer.
In the exercise, we identified using the substitution \(u = \sin x\), which made the integral \(\int e^{u} \, du\) much simpler to solve. After integrating and finding the antiderivative, we substitute back to get \(F(x) = e^{\sin x} + C\).
Chain Rule
The chain rule in calculus is a formula for computing the derivative of the composition of two functions. It states that if you have a function \(g(x)\) inside another function \(f(u)\), where \(u = g(x)\), then the derivative of the composite function is \(f'(g(x)) \, g'(x)\).During the verification step of finding the antiderivative for \(f(x) = e^{\sin x} \cos x\), we used the chain rule to differentiate \(F(x) = e^{\sin x} + C\). Here are the steps:- Identify \(g(x) = \sin x\) inside \(F(x) = e^{\sin x}\).- Differentiate the outer function \(e^u\) with respect to \(u\) and get \(e^u\).- Multiply by the derivative of the inner function \(\sin x\), which is \(\cos x\).Thus, \(\frac{d}{dx}[e^{\sin x} + C] = e^{\sin x} \cdot \cos x = f(x)\). The chain rule confirms that our antiderivative is correct.

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