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Given information
The given equation is
Step-by-step explanation
Step 1.
We will use Integration by Parts.
\begin{aligned} v^{\prime} &=& \sin\;4 x \; \\\\ \therefore \;\; v &=& - \frac {1}{4} \; \cos\;4 x \\\\ \therefore \;\; \int \; e^{3 x} \; \cos\;4 x \; dx &=& \frac {e^{3 x} \; \sin\;4 x}{4} - \frac {3}{4} \; [\; e^{3 x} \cdot (- \frac {1}{4} \; \cos\;4 x) - \int\; 3 \; e^{3 x} \cdot (- \frac {1}{4} \; \cos\;4 x) \;\;d x \;] \\\\ &=& \frac {e^{3 x} \; \sin\;4 x}{4} + \frac {3}{16} \; e^{3 x} \cdot \cos\;4 x - \frac {3}{4} \cdot \frac {3}{4} \; \int\; e^{3 x} \; \cos\;4 x \;\; d x \;] \\\\ \therefore \;\; \int \; e^{3 x} \; \cos\;4 x \; dx &=& \frac {e^{3 x} \; \sin\;4 x}{4} + \frac {3}{16} \; e^{3 x} \cdot \cos\;4 x - \frac {9}{16} \; \int\; e^{3 x} \; \cos\;4 x \;\;d x \\\\ \int \; e^{3 x} \; \cos\;4 x \; dx + \frac {9}{16} \; \int\; e^{3 x} \; \cos\;4 x \; d x &=& \frac {e^{3 x} \; \sin\;4 x}{4} + \frac {3}{16} \; e^{3 x} \cdot \cos\;4 x \\\\ \frac {25}{16} \; \int\; e^{3 x} \; \cos\;4 x \;\; d x &=& \frac {e^{3 x} \; \sin\;4 x}{4} + \frac {3}{16} \; e^{3 x} \cdot \cos\;4 x \\\\ \frac {25}{4} \; \int\; e^{3 x} \; \cos\;4 x \;\; d x &=& e^{3 x} \; \sin\;4 x + \frac {3}{4} \; e^{3 x} \cdot \cos\;4 x \\\\ \int\; e^{3 x} \; \cos\;4 x \;\; d x &=& \textcolor{#000764}{ \frac {4}{25} \;e^{3 x} \; \sin\;4 x + \frac {3}{25} \; e^{3 x} \; \cos\;4 x + C } \\\\ \end{aligned}
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CALCULUS:EARLY TRANSCENDENTALS
4 Edition,
Rogawski
ISBN: 9781319050740
Solutions
Chapter
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- Problem 3
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- Problem 50a
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- Problem 70
- Problem 71
- Problem 72a
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- Problem 93a
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- Problem 96a
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- Problem 96c
- Problem 96d
- Problem 96e
- Problem 97a
- Problem 97b
- Problem 97c