Question Statement

Evaluate the integral:

$\int e^x\cos xdx$

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Background and Explanation

This integral requires integration by parts twice. When integrating products of exponential and trigonometric functions, the original integral typically reappears after two applications, allowing you to solve for it algebraically—a technique sometimes called "integration by parts with cyclic repetition."

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Solution

Let us denote the integral by $I$:

$I=\int e^x\cos xdx$

First Application of Integration by Parts

We apply integration by parts using the formula $\int udv=uv-\int vdu$.

Choose:

• $u=\cos x$ (so that $du=-\sin xdx$)
• $dv=e^{x}dx$ (so that $v=e^x$)

Applying the formula:

$\begin{array}{rl}I& =\cos x\cdot \int e^{x}dx-\int \left(\int e^{x}dx\right)\cdot \frac{d}{dx}(\cos x)dx\\ & =\cos x\cdot e^x-\int e^x\cdot (-\sin x)dx\\ & =e^x\cos x+\int e^x\sin xdx\end{array}$

Second Application of Integration by Parts

The new integral $\int e^x\sin xdx$ also requires integration by parts. Choose:

• $u=\sin x$ (so that $du=\cos xdx$)
• $dv=e^{x}dx$ (so that $v=e^x$)

Substituting back:

$\begin{array}{rl}I& =e^x\cos x+\left[\sin x\cdot \int e^{x}dx-\int \left(\int e^{x}dx\right)\cdot \frac{d}{dx}(\sin x)dx\right]\\ & =e^x\cos x+\sin x\cdot e^x-\int e^x\cos xdx\end{array}$

Solving for $I$

Notice that the original integral $I=\int e^x\cos xdx$ has reappeared on the right-hand side:

$I=e^x\cos x+e^x\sin x-I$

Now we solve this equation for $I$:

$\begin{array}{rl}I+I& =e^x\cos x+e^x\sin x+C\\ 2I& =e^x(\cos x+\sin x)+C\\ I& =\frac{e^x}{2}(\cos x+\sin x)+c\end{array}$

(Note: We add the constant of integration $C$ when integrating, which becomes $c=C/2$ after dividing by 2.)

Therefore, the final solution is:

$\int e^x\cos xdx=\frac{e^x}{2}(\cos x+\sin x)+c$

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Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• LIATE Rule (for choosing $u$): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—here we chose trigonometric functions as $u$ and exponential as $dv$
• Cyclic Integration Technique: When the original integral reappears after integration by parts, treat it as an algebraic variable and solve for it

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Summary of Steps

1. Define $I=\int e^x\cos xdx$ to enable algebraic manipulation later
2. First integration by parts: Set $u=\cos x$ and $dv=e^{x}dx$ to obtain $I=e^x\cos x+\int e^x\sin xdx$
3. Second integration by parts: Apply the method to $\int e^x\sin xdx$ with $u=\sin x$ and $dv=e^{x}dx$
4. Identify the cycle: The original integral $I$ reappears on the right side, giving you $I=e^x(\cos x+\sin x)-I$
5. Solve algebraically: Add $I$ to both sides to get $2I=e^x(\cos x+\sin x)+C$
6. Isolate the answer: Divide by 2 to obtain $I=\frac{e^x}{2}(\cos x+\sin x)+c$