Integrate with respect to x x x ∫ e 5 x sin 3 x d x \int e^{5x} \sin 3x \, dx ∫ e 5 x sin 3 x d x
a. 3 34 e 5 x ( 1 3 sin 3 x − cos 3 x ) + c \frac{3}{34} e^{5x}\left(\frac{1}{3} \sin 3x - \cos 3x\right) + c 34 3 e 5 x ( 3 1 sin 3 x − cos 3 x ) + c
b. 3 24 e 5 x ( 5 3 sin 3 x − cos 3 x ) + c \frac{3}{24} e^{5x}\left(\frac{5}{3} \sin 3x - \cos 3x\right) + c 24 3 e 5 x ( 3 5 sin 3 x − cos 3 x ) + c
c. 3 34 e 5 x ( 5 3 sin 3 x − cos 3 x ) + c − yes ∗ \frac{3}{34} e^{5x}\left(\frac{5}{3} \sin 3x - \cos 3x\right) + c \quad -\quad \text{yes}^* 34 3 e 5 x ( 3 5 sin 3 x − cos 3 x ) + c − yes ∗
d. 3 34 e 5 x ( 1 3 sin 3 x − cos 2 x ) + c \frac{3}{34} e^{5x}\left(\frac{1}{3} \sin 3x - \cos 2x\right) + c 34 3 e 5 x ( 3 1 sin 3 x − cos 2 x ) + c
Solution.
We must integrate it twice by parts using this formula (or theorem):
∫ u d v = v u − ∫ v d u \int u \, dv = vu - \int v \, du ∫ u d v = vu − ∫ v d u
Let's find this integral:
I = ∫ e 5 x sin 3 x d x = ∣ u = sin 3 x d u = d sin 3 x = 3 cos 3 x d x d v = e 5 x d x v = ∫ e 5 x d x = 1 5 e 5 x ∣ = 1 5 e 5 x sin 3 x − ∫ 1 5 e 5 x ⋅ 3 cos 3 x d x = = 1 5 e 5 x sin 3 x − 3 5 ∫ e 5 x cos 3 x d x = 1 5 e 5 x sin 3 x − 3 5 I 1 \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} I = ∫ e 5 x sin 3 x d x = ∣ ∣ u = sin 3 x d v = e 5 x d x d u = d sin 3 x = 3 cos 3 x d x v = ∫ e 5 x d x = 5 1 e 5 x ∣ ∣ = 5 1 e 5 x sin 3 x − ∫ 5 1 e 5 x ⋅ 3 cos 3 x d x = = 5 1 e 5 x sin 3 x − 5 3 ∫ e 5 x cos 3 x d x = 5 1 e 5 x sin 3 x − 5 3 I 1
Let's find I 1 I_1 I 1
∫ e 5 x cos 3 x d x = ∣ u = cos 3 x d u = d cos 3 x = − 3 sin 3 x d x d v = e 5 x d x v = ∫ e 5 x d x = 1 5 e 5 x ∣ = 1 5 e 5 x cos 3 x − ∫ 1 5 e 5 x ⋅ ( − 3 sin 3 x ) d x = = 1 5 e 5 x cos 3 x + 3 5 ∫ e 5 x sin 3 x d x \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} ∫ e 5 x cos 3 x d x = ∣ ∣ u = cos 3 x d v = e 5 x d x d u = d cos 3 x = − 3 sin 3 x d x v = ∫ e 5 x d x = 5 1 e 5 x ∣ ∣ = 5 1 e 5 x cos 3 x − ∫ 5 1 e 5 x ⋅ ( − 3 sin 3 x ) d x = = 5 1 e 5 x cos 3 x + 5 3 ∫ e 5 x sin 3 x d x
So
I = 1 5 e 5 x sin 3 x − 3 5 I 1 = 1 5 e 5 x sin 3 x − 3 5 ( 1 5 e 5 x cos 3 x + 3 5 ∫ e 5 x sin 3 x d x ) = 1 5 e 5 x sin 3 x − 3 25 e 5 x cos 3 x − − 9 25 ∫ e 5 x sin 3 x d x = 1 5 e 5 x sin 3 x − 3 25 e 5 x cos 3 x − 9 25 I = I \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} I = 5 1 e 5 x sin 3 x − 5 3 I 1 = 5 1 e 5 x sin 3 x − 5 3 ( 5 1 e 5 x cos 3 x + 5 3 ∫ e 5 x sin 3 x d x ) = 5 1 e 5 x sin 3 x − 25 3 e 5 x cos 3 x − − 25 9 ∫ e 5 x sin 3 x d x = 5 1 e 5 x sin 3 x − 25 3 e 5 x cos 3 x − 25 9 I = I
And now we can find I I I
1 5 e 5 x sin 3 x − 3 25 e 5 x cos 3 x − 9 25 I = I I ( 1 + 9 25 ) = e 5 x ( 1 5 sin 3 x − 3 25 cos 3 x ) 34 25 I = e 5 x ⋅ 5 sin 3 x − 3 cos 3 x 25 ∣ × 25 \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} 5 1 e 5 x sin 3 x − 25 3 e 5 x cos 3 x − 25 9 I = I I ( 1 + 25 9 ) = e 5 x ( 5 1 sin 3 x − 25 3 cos 3 x ) 25 34 I = e 5 x ⋅ 25 5 s i n 3 x − 3 c o s 3 x ∣ × 25 I = e 5 x 34 ( 5 sin 3 x − 3 cos 3 x ) + c I = \frac{e^{5x}}{34} (5 \sin 3x - 3 \cos 3x) + c I = 34 e 5 x ( 5 sin 3 x − 3 cos 3 x ) + c I = 3 34 e 5 x ( 5 3 sin 3 x − cos 3 x ) + c I = \frac{3}{34} e^{5x} \left( \frac{5}{3} \sin 3x - \cos 3x \right) + c I = 34 3 e 5 x ( 3 5 sin 3 x − cos 3 x ) + c
Answer: variant c is correct:
c . 3 34 e 5 x ( 5 3 sin 3 x − cos 3 x ) + c \mathbf{c}. \frac{3}{34} e^{5x} \left( \frac{5}{3} \sin 3x - \cos 3x \right) + c c . 34 3 e 5 x ( 3 5 sin 3 x − cos 3 x ) + c 
#340153 on Dec 2023