First slide

Introduction to integration

Question

If $\int e^{x} \cos 4 x d x = A e^{5 x} (\sin 4 x + \frac{5}{4} \cos 4 x) + C$
then $A$ is equal to

Easy

A

$\frac{4}{41}$
B

$\frac{3}{41}$
C

$\frac{5}{41}$
D

$\frac{9}{41}$

Solution

Integrating by parts taking $e^{4 x}$ as first function, we have
$\begin{array}{l} I = \int e^{5 x} \cos 4 x d x = \frac{1}{4} e^{5 x} \sin 4 x - \frac{5}{4} \int e^{5 x} \sin 4 x d x \\ = \frac{1}{4} e^{5 x} \sin 4 x - \frac{5}{4} [- e^{5 x} \frac{\cos 4 x}{4} + \frac{5}{4} \int e^{5 x} \cos 4 x d x] \\ = \frac{1}{4} e^{5 x} \sin 4 x + \frac{5}{16} e^{5 x} \cos 4 x - \frac{25}{16} I \\ \Rightarrow (1 + \frac{25}{16}) I = \frac{4}{16} e^{5 x} (\sin 4 x + \frac{5}{4} \cos 4 x) \\ \Rightarrow I = \frac{4}{41} e^{5 x} (\sin 4 x + \frac{5}{4} \cos x) + C \end{array}$

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