int e^5xsin3xdx =
∫ e^(5x) sin(3x) dx
=(i/2)∫ e^(5x) (e^(-3ix)-e^(3ix)) dx
=(i/2)∫ e^((5-3i)x)-e^((5+3i)x) dx
=(i/2)((5+3i)e^((5-3i)x)/34-(5-3i)e^((5+3i)x)/34) + C
= (5i/78)e^(5x)(e^(-3ix)-e^(3ix))+(-3/78)e^(5x)(e^(-3ix)+e^(3ix)) + C
= (5/34)e^(5x)sin(3x) - (3/34)e^(5x)cos(3x)
= e^(5x)(5sin(3x)-3cos(5x))/34
To solve the integral of e^(5x) sin(3x) dx, we will use integration by parts.
Let's consider the integral of u dv = uv - integral of v du, where u and v are functions.
In this case, we can choose u = sin(3x) and dv = e^(5x) dx.
First, let's find the derivatives of u and integrals of dv:
du/dx = 3 cos(3x)
v =∫e^(5x) dx
Now, we need to find v:
To find the integral of e^(5x) dx, we can use the power rule. The integral of e^(kx) dx = e^(kx)/k, so in this case:
v = ∫e^(5x) dx = e^(5x)/5
Next, we substitute these values into the integration by parts formula:
∫e^(5x) sin(3x) dx = uv - ∫v du
∫e^(5x) sin(3x) dx = sin(3x) * (e^(5x)/5) - ∫(e^(5x)/5) * (3 cos(3x)) dx
We simplify:
∫e^(5x) sin(3x) dx = (e^(5x)/5) * sin(3x) - 3/5 * ∫e^(5x) cos(3x) dx
Now, we have another integral to solve, namely ∫e^(5x) cos(3x) dx. We can use the same method and integration by parts again:
Let u = cos(3x) and dv = e^(5x) dx
du/dx = -3 sin(3x)
v = ∫e^(5x) dx = e^(5x)/5
Substituting these values into the integration by parts formula:
∫e^(5x) cos(3x) dx = uv - ∫v du
∫e^(5x) cos(3x) dx = cos(3x) * (e^(5x)/5) - ∫(e^(5x)/5) * (-3 sin(3x)) dx
Simplifying:
∫e^(5x) cos(3x) dx = (e^(5x)/5) * cos(3x) + 3/5 * ∫e^(5x) sin(3x) dx
Now, let's substitute this back into the original equation:
∫e^(5x) sin(3x) dx = (e^(5x)/5) * sin(3x) - 3/5 * ((e^(5x)/5) * cos(3x) + 3/5 * ∫e^(5x) sin(3x) dx)
We can rearrange this equation to solve for ∫e^(5x) sin(3x) dx:
∫e^(5x) sin(3x) dx = (e^(5x)/5) * sin(3x) - (9/25) * (e^(5x)/5) * cos(3x)
Simplifying further:
∫e^(5x) sin(3x) dx = (1/5) * e^(5x) * (sin(3x) - 9cos(3x)) + C
So, the final solution is ∫e^(5x) sin(3x) dx = (1/5) * e^(5x) * (sin(3x) - 9cos(3x)) + C, where C is the constant of integration.
To compute the integral of e^5x * sin(3x) with respect to x, we can use the technique of integration by parts. The formula for integration by parts is given as follows:
∫ u * dv = uv - ∫ v * du
Let's assign u and dv to our function:
u = sin(3x) (and hence, du = 3cos(3x)dx)
dv = e^5x dx (and hence, v = (1/5)e^5x)
Using the formula for integration by parts, we can rewrite the integral as:
∫ e^5x * sin(3x)dx = (1/5)e^5x * sin(3x) - ∫ (1/5)e^5x * 3cos(3x) dx
Simplifying this expression further, we get:
∫ e^5x * sin(3x) dx = (1/5)e^5x * sin(3x) - (3/5) ∫ e^5x * cos(3x) dx
Now we have a new integral to compute, namely:
∫ e^5x * cos(3x) dx
To solve this new integral, we can again use integration by parts. Assigning u and dv as follows:
u = cos(3x) (and hence, du = -3sin(3x)dx)
dv = e^5x dx (and hence, v = (1/5)e^5x)
Using the formula for integration by parts again, we can rewrite the integral as:
∫ e^5x * cos(3x) dx = (1/5)e^5x * cos(3x) - ∫ (1/5)e^5x * -3sin(3x) dx
Simplifying this expression further, we get:
∫ e^5x * cos(3x) dx = (1/5)e^5x * cos(3x) + (3/5) ∫ e^5x * sin(3x) dx
Notice that the integral on the right side of the equation is the original integral we wanted to compute. So, we can substitute it back into the equation:
∫ e^5x * sin(3x) dx = (1/5)e^5x * sin(3x) - (3/5) * [ (1/5)e^5x * cos(3x) + (3/5) ∫ e^5x * sin(3x) dx ]
Now, let's simplify this equation and solve for the integral that we want:
∫ e^5x * sin(3x) dx = (1/5)e^5x * sin(3x) - (3/25)e^5x * cos(3x) - (9/25) ∫ e^5x * sin(3x) dx
To isolate the ∫ e^5x * sin(3x) dx term on one side, we can rearrange the equation:
(34/25) ∫ e^5x * sin(3x) dx = (1/5)e^5x * sin(3x) - (3/25)e^5x * cos(3x)
Finally, we can solve for the integral by dividing both sides by (34/25):
∫ e^5x * sin(3x) dx = (5/34)e^5x * sin(3x) - (15/34)e^5x * cos(3x) + C
Therefore, the value of the integral is given by:
∫ e^5x * sin(3x) dx = (5/34)e^5x * sin(3x) - (15/34)e^5x * cos(3x) + C
where C represents the constant of integration.