Evaluate $$ \int e^{x} \sin x d x $$

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$$ \int e^{x} \sin x \, dx = \frac{e^{x}}{2} (\sin x - \cos x) + C $$ To evaluate the integral $$ \int e^{x} \sin x \, dx $$, we can use the method of integration by parts or the technique of integration by parts twice. However, a more systematic approach is to use the formula for integrating products of exponentials and trigonometric functions. We will use the formula: $$ \int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C $$ In our case, we have $$ a = 1 $$ and $$ b = 1 $$. Thus, we can substitute these values into the formula: $$ \int e^{x} \sin x \, dx = \frac{e^{x}}{1^2 + 1^2} (1 \sin x - 1 \cos x) + C $$ Calculating $$ 1^2 + 1^2 $$: $$ 1^2 + 1^2 = 1 + 1 = 2 $$ Now substituting this back into the formula gives: $$ \int e^{x} \sin x \, dx = \frac{e^{x}}{2} (\sin x - \cos x) + C $$ Thus, the final result is: $$ \int e^{x} \sin x \, dx = \frac{e^{x}}{2} (\sin x - \cos x) + C $$

Evaluate \( \int e^{x} \sin x d x \)

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Evaluate \( \int e^{x} \sin x d x \)

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