Question

Integrate with respect to x: ∫ e^5x sin⁡ 3x dx

a. 3e^5x/34(1/3 sin⁡ 3x-cos⁡ 3x)+c

b. 3e^5x/24(5/3 sin 3x-cos⁡ 3x)+c

c. 3e^5x/34(5/3 sin⁡ 3x-cos⁡ 3x)+c - yes* (please let me know if this is correct)

d. 3e^5x/34(1/3 sin⁡ 3x-cos⁡ 2x)+c

Answer 1

Integrate with respect to $x$: $\int e^{5x} \sin 3x \, dx$

a. $\frac{3}{34} e^{5x}\left(\frac{1}{3} \sin 3x - \cos 3x\right) + c$

b. $\frac{3}{24} e^{5x}\left(\frac{5}{3} \sin 3x - \cos 3x\right) + c$

c. $\frac{3}{34} e^{5x}\left(\frac{5}{3} \sin 3x - \cos 3x\right) + c \quad -\quad \text{yes}^*$ (please let me know if this is correct)

d. $\frac{3}{34} e^{5x}\left(\frac{1}{3} \sin 3x - \cos 2x\right) + c$

Solution.

We must integrate it twice by parts using this formula (or theorem):

$$\int u \, dv = vu - \int v \, du$$

Let's find this integral:

$$\begin{array}{l} I = \int e^{5x} \sin 3x \, dx = \left| \begin{array}{cc} u = \sin 3x & du = d \sin 3x = 3 \cos 3x \, dx \\ dv = e^{5x} \, dx & v = \int e^{5x} \, dx = \frac{1}{5} e^{5x} \end{array} \right| = \frac{1}{5} e^{5x} \sin 3x - \int \frac{1}{5} e^{5x} \cdot 3 \cos 3x \, dx = \\ = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{5} \int e^{5x} \cos 3x \, dx = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{5} I_1 \end{array}$$

Let's find $I_1$:

$$\begin{array}{l} \int e^{5x} \cos 3x \, dx = \left| \begin{array}{cc} u = \cos 3x & du = d \cos 3x = -3 \sin 3x \, dx \\ dv = e^{5x} \, dx & v = \int e^{5x} \, dx = \frac{1}{5} e^{5x} \end{array} \right| = \frac{1}{5} e^{5x} \cos 3x - \int \frac{1}{5} e^{5x} \cdot (-3 \sin 3x) \, dx = \\ = \frac{1}{5} e^{5x} \cos 3x + \frac{3}{5} \int e^{5x} \sin 3x \, dx \end{array}$$

So

$$\begin{array}{l} I = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{5} I_1 = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{5} \left(\frac{1}{5} e^{5x} \cos 3x + \frac{3}{5} \int e^{5x} \sin 3x \, dx\right) = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{25} e^{5x} \cos 3x - \\ - \frac{9}{25} \int e^{5x} \sin 3x \, dx = \frac{1}{5} e^{5x} \sin 3x - \frac{3}{25} e^{5x} \cos 3x - \frac{9}{25} I = I \end{array}$$

And now we can find $I$ from the equation:

$$\begin{array}{l} \frac{1}{5} e^{5x} \sin 3x - \frac{3}{25} e^{5x} \cos 3x - \frac{9}{25} I = I \\ I \left(1 + \frac{9}{25}\right) = e^{5x} \left(\frac{1}{5} \sin 3x - \frac{3}{25} \cos 3x\right) \\ \frac{34}{25} I = e^{5x} \cdot \frac{5 \sin 3x - 3 \cos 3x}{25} \quad \mid \times 25 \end{array}$$ $$I = \frac{e^{5x}}{34} (5 \sin 3x - 3 \cos 3x) + c$$ $$I = \frac{3}{34} e^{5x} \left( \frac{5}{3} \sin 3x - \cos 3x \right) + c$$

Answer: variant c is correct:

$$\mathbf{c}. \frac{3}{34} e^{5x} \left( \frac{5}{3} \sin 3x - \cos 3x \right) + c$$