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hello,

 

I'm trying to find the indefinite integral of :

 

\(\int e^{5x}sin(3x) dx\)

 

I get to a point and it just becomes a loop?

 

where do I go from here?

 

I get to = \(\frac{e^{5x}sin(3x)}{5}-\frac{3e^{5x}cos(3x)}{25}-\int \frac{-9e^{5x}sin(3x)}{25}dx\)

then solving for \(\int \frac{-9e^{5x}sin(3x)}{25}dx\)

 

becomes: \(\frac{-9}{25}\int e^{5x}sin(3x)\)

 

then the loop repeats, I'm so confused where to go from here.

 

Your help is much appreciated !

vest4R  May 27, 2017
 #1
avatar
+2

Take the integral:
 integral e^(5 x) sin(3 x) dx


For the integrand e^(5 x) sin(3 x), integrate by parts, integral f dg = f g - integral g df, where 
 f = sin(3 x), dg = e^(5 x) dx, df = 3 cos(3 x) dx, g = e^(5 x)/5:


 integral e^(5 x) sin(3 x) dx = 1/5 e^(5 x) sin(3 x) - 3/5 integral e^(5 x) cos(3 x) dx
For the integrand e^(5 x) cos(3 x), integrate by parts, integral f dg = f g - integral g df, where 
 f = cos(3 x), dg = e^(5 x) dx, df = -3 sin(3 x) dx, g = e^(5 x)/5:


 integral e^(5 x) sin(3 x) dx = 1/5 e^(5 x) sin(3 x) - 3/25 e^(5 x) cos(3 x) - 9/25 integral e^(5 x) sin(3 x) dx
Add 9/25 integral e^(5 x) sin(3 x) dx to both sides:


34/25 integral e^(5 x) sin(3 x) dx = 1/5 e^(5 x) sin(3 x) - 3/25 e^(5 x) cos(3 x) + constant
Multiply both sides by 25/34:


 integral e^(5 x) sin(3 x) dx = 25/34 (1/5 e^(5 x) sin(3 x) - 3/25 e^(5 x) cos(3 x)) + constant
Which is equal to:
Answer: | = 1/34 e^(5 x) (5 sin(3 x) - 3 cos(3 x)) + constant

Guest May 27, 2017
 #2
avatar+93866 
+3

Yes, there is a loop :)

 

\(\int e^{5x}sin(3x)dx=[\frac{e^{5x}}{5}\cdot sin(3x)]-\int \frac{e^{5x}}{5}\cdot3cos(3x)dx\\ \int e^{5x}sin(3x)dx=\frac{e^{5x} sin(3x)}{5}-\frac{3}{5}\int e^{5x}cos(3x)dx\\ \int e^{5x}sin(3x)dx=\frac{e^{5x} sin(3x)}{5}-\frac{3}{5}\left[(\frac{e^{5x}cos(3x)}{5})-\int \frac{-3}{5} e^{5x}sin(3x)dx\right]\\ \int e^{5x}sin(3x)dx=\frac{e^{5x} sin(3x)}{5}-\frac{3e^{5x}cos(3x)}{25}- \frac{9}{25}\int e^{5x}sin(3x)dx+c\\ \int e^{5x}sin(3x)dx+ \frac{9}{25}\int e^{5x}sin(3x)dx=\frac{e^{5x} sin(3x)}{5}-\frac{3e^{5x}cos(3x)}{25}+c\\ \frac{34}{25}\int e^{5x}sin(3x)dx=\frac{5e^{5x} sin(3x)}{25}-\frac{3e^{5x}cos(3x)}{25}+c\\ \int e^{5x}sin(3x)dx=\frac{5e^{5x} sin(3x)}{34}-\frac{3e^{5x}cos(3x)}{34}+c\\ \int e^{5x}sin(3x)dx=\frac{e^{5x} (5sin(3x)-3cos(3x))}{34}+c \)

Melody  May 27, 2017

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