What is \(\smallint {{\rm{e}}^{\sin {\rm{x}}}}\left( {\frac{{{\rm{x}}{{\cos }^3}{\rm{x}} - \sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}} \right){\rm{dx}}\) equal to?
(x – sec x)e sin x + c
Evaluating Complex Integrals: A Step-by-Step Solution
We are asked to evaluate the indefinite integral: \(\smallint {{\rm{e}}^{\sin {\rm{x}}}}\left( {\frac{{{\rm{x}}{{\cos }^3}{\rm{x}} - \sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}} \right){\rm{dx}}\)
Simplifying the Integrand Expression
The integrand involves a fraction inside the parenthesis. Let's simplify this fraction by splitting it into two terms:
\(\frac{{{\rm{x}}{{\cos }^3}{\rm{x}} - \sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}} = \frac{{{\rm{x}}{{\cos }^3}{\rm{x}}}}{{{{\cos }^2}{\rm{x}}}} - \frac{{\sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}\)
Simplifying each term:
- The first term: \(\frac{{{\rm{x}}{{\cos }^3}{\rm{x}}}}{{{{\cos }^2}{\rm{x}}}} = {\rm{x}}\cos {\rm{x}}\)
- The second term: \(\frac{{\sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}} = \frac{{\sin {\rm{x}}}}{{\cos {\rm{x}}}} \cdot \frac{1}{{\cos {\rm{x}}}} = \tan {\rm{x}}\sec {\rm{x}}\)
So, the expression inside the parenthesis simplifies to: \({\rm{x}}\cos {\rm{x}} - \tan {\rm{x}}\sec {\rm{x}}\)
Alternatively, the second term can also be written as \(\sin {\rm{x}} \cdot \frac{1}{{{\cos }^2}{\rm{x}}} = \sin {\rm{x}}{\sec ^2}{\rm{x}}\). So the simplified expression is \({\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}\).
Thus, the integral becomes: \(\smallint {{\rm{e}}^{\sin {\rm{x}}}}\left( {{\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}} \right){\rm{dx}}\)
Recognizing a Standard Integration Form
This integral has the form of \({\rm{e}}^{f({\rm{x}})}\) multiplied by an expression involving \({\rm{x}}\) and trigonometric functions. A common integration formula involving \({\rm{e}}^{f({\rm{x}})}\) is:
\(\smallint {{\rm{e}}^{f({\rm{x}})}}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right){\rm{dx}} = {{\rm{e}}^{f({\rm{x}})}}g({\rm{x}}) + C\)
Let's see if our integral fits this form.
In our integral, we have \({\rm{e}}^{\sin {\rm{x}}}\), which suggests that \(f({\rm{x}}) = \sin {\rm{x}}\). The derivative of \(f({\rm{x}})\) is \(f'({\rm{x}}) = \cos {\rm{x}}\).
So, we want to see if the expression \(\left( {{\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}} \right)\) can be written in the form \(g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})\), which is \(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}})\).
We need to find a function \(g({\rm{x}})\) such that:
\(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}}) = {\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}\)
Finding the Function g(x)
Let's look at the options provided, as they suggest the form of \(g({\rm{x}})\).
The options suggest \(g({\rm{x}})\) could be of the form \({\rm{x}} \pm \sec {\rm{x}}\) or \({\rm{x}} \pm \tan {\rm{x}}\).
Let's test the option that corresponds to the correct answer:
Suppose \(g({\rm{x}}) = {\rm{x}} - \sec {\rm{x}}\).
First, find the derivative of \(g({\rm{x}})\):
\({g}'({\rm{x}}) = \frac{d}{{d{\rm{x}}}}({\rm{x}} - \sec {\rm{x}}) = \frac{d}{{d{\rm{x}}}}({\rm{x}}) - \frac{d}{{d{\rm{x}}}}(\sec {\rm{x}}) = 1 - \sec {\rm{x}}\tan {\rm{x}}\)
Now, let's calculate \(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}})\) using this \(g({\rm{x}})\) and \(g'({\rm{x}})\):
\(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}}) = ({\rm{x}} - \sec {\rm{x}})\cos {\rm{x}} + (1 - \sec {\rm{x}}\tan {\rm{x}})\)
Expand the first term:
\(({\rm{x}} - \sec {\rm{x}})\cos {\rm{x}} = {\rm{x}}\cos {\rm{x}} - \sec {\rm{x}}\cos {\rm{x}}\)
Recall that \(\sec {\rm{x}} = \frac{1}{{\cos {\rm{x}}}}\). So, \(\sec {\rm{x}}\cos {\rm{x}} = \frac{1}{{\cos {\rm{x}}}}\cos {\rm{x}} = 1\).
Thus, \(({\rm{x}} - \sec {\rm{x}})\cos {\rm{x}} = {\rm{x}}\cos {\rm{x}} - 1\).
Now substitute this back into the expression for \(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}})\):
\(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}}) = ({\rm{x}}\cos {\rm{x}} - 1) + (1 - \sec {\rm{x}}\tan {\rm{x}})\)
\(= {\rm{x}}\cos {\rm{x}} - 1 + 1 - \sec {\rm{x}}\tan {\rm{x}}\)
\(= {\rm{x}}\cos {\rm{x}} - \sec {\rm{x}}\tan {\rm{x}}\)
Recall that \(\sec {\rm{x}}\tan {\rm{x}} = \frac{1}{{\cos {\rm{x}}}} \cdot \frac{{\sin {\rm{x}}}}{{\cos {\rm{x}}}} = \frac{{\sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}} = \sin {\rm{x}}{\sec ^2}{\rm{x}}\).
So, \(g({\rm{x}})\cos {\rm{x}} + {g}'({\rm{x}}) = {\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}\).
This matches the simplified integrand we found earlier: \(\left( {{\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}} \right)\).
Therefore, the integrand is indeed in the form \({\rm{e}}^{f({\rm{x}})}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right)\) with \(f({\rm{x}}) = \sin {\rm{x}}\) and \(g({\rm{x}}) = {\rm{x}} - \sec {\rm{x}}\).
Applying the Integration Formula
Using the formula \(\smallint {{\rm{e}}^{f({\rm{x}})}}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right){\rm{dx}} = {{\rm{e}}^{f({\rm{x}})}}g({\rm{x}}) + C\), we have:
\(\smallint {{\rm{e}}^{\sin {\rm{x}}}}\left( {{\rm{x}}\cos {\rm{x}} - \sin {\rm{x}}{\sec ^2}{\rm{x}}} \right){\rm{dx}} = {{\rm{e}}^{\sin {\rm{x}}}}({\rm{x}} - \sec {\rm{x}}) + C\)
Final Result
The integral is equal to \(({\rm{x}} - \sec {\rm{x}}){{\rm{e}}^{\sin {\rm{x}}}} + C\).
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Integral Evaluation | The process of finding the antiderivative of a function. |
| Simplifying Integrands | Algebraic manipulation of the function being integrated to make it easier to integrate. |
| Standard Integral Form | Recognizing specific patterns in integrals that match known formulas, such as \(\smallint {{\rm{e}}^{f({\rm{x}})}}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right){\rm{dx}}\). |
| Differentiation Rules | Necessary for verifying the parts of the standard integral form, e.g., the derivative of \(\sec {\rm{x}}\). |
Additional Information on Integration Techniques
This problem highlights how recognizing specific forms can simplify complex integrals. While techniques like integration by parts are powerful, sometimes algebraic manipulation and pattern recognition lead directly to the solution. The key form used here, \(\smallint {{\rm{e}}^{f({\rm{x}})}}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right){\rm{dx}} = {{\rm{e}}^{f({\rm{x}})}}g({\rm{x}}) + C\), is derived from the product rule for differentiation. If you differentiate \({\rm{e}}^{f({\rm{x}})}g({\rm{x}})\) using the product rule and chain rule, you get:
\(\frac{d}{{d{\rm{x}}}}\left( {{{\rm{e}}^{f({\rm{x}})}}g({\rm{x}})} \right) = \frac{d}{{d{\rm{x}}}}\left( {{{\rm{e}}^{f({\rm{x}})}}} \right)g({\rm{x}}) + {{\rm{e}}^{f({\rm{x}})}}\frac{d}{{d{\rm{x}}}}\left( {g({\rm{x}})} \right)\)
\(= {{\rm{e}}^{f({\rm{x}})}}{f}'({\rm{x}})g({\rm{x}}) + {{\rm{e}}^{f({\rm{x}})}}{g}'({\rm{x}}))\)
\(= {{\rm{e}}^{f({\rm{x}})}}\left( {g({\rm{x}}){f}'({\rm{x}}) + {g}'({\rm{x}})} \right)\)
Integrating both sides gives the formula. This technique is particularly useful when the integrand is a product of \({\rm{e}}^{f({\rm{x}})}\) and a sum where one term contains \(f'({\rm{x}})\) multiplied by some function \(g({\rm{x}})\), and the other term is the derivative of that function \(g'({\rm{x}})\).
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