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integration of e^{ax} sin bx

The integral ($\int e^{ax} \sin(bx) \, dx$) is computed as:

\int e^{ax} \sin(bx) \, dx= \frac{a e^{ax} \sin(bx)}{a^2 + b^2} – \frac{b e^{ax} \cos(bx)}{a^2 + b^2}+ C

Proof of Integration

To solve the integral $\int e^{ax} \sin(bx) \, dx$, we can use the method of integration by parts, which is based on the formula:

\int u \, dv = uv – \int v \, du

Since the integral involves a product of two functions ($e^{ax}$ and $\sin(bx)$), and because direct integration isn’t straightforward, integration by parts is a suitable approach. In this case, we’ll apply the method twice, because the first application will bring us back to a similar integral form, allowing us to solve for the integral in terms of itself.

Let’s set:

  • $u = e^{ax}$, which implies $du = a e^{ax} dx$
  • $dv = \sin(bx) dx$, which implies $v = -\frac{1}{b} \cos(bx)$

Applying integration by parts, we get:

\int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) – \int -\frac{a}{b} e^{ax} \cos(bx) \, dx

For the second integral, we apply integration by parts again, this time with:

  • $u = e^{ax}$ (again), resulting in $du = a e^{ax} dx$
  • $dv = \cos(bx) dx$, resulting in $v = \frac{1}{b} \sin(bx)$

Thus, the integral becomes:

-\frac{a}{b} \int e^{ax} \cos(bx) \, dx = -\frac{a}{b^2} e^{ax} \sin(bx) + \frac{a^2}{b^2} \int e^{ax} \sin(bx) \, dx

Putting it all together, we have:

\int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) + \frac{a}{b^2} e^{ax} \sin(bx) – \frac{a^2}{b^2} \int e^{ax} \sin(bx) \, dx

Solving for the integral, we get:

\left(1 + \frac{a^2}{b^2}\right) \int e^{ax} \sin(bx) \, dx = -\frac{e^{ax}}{b} \cos(bx) + \frac{a}{b^2} e^{ax} \sin(bx)

Therefore, the integral is:

\int e^{ax} \sin(bx) \, dx = \frac{-e^{ax} \cos(bx)}{b + \frac{a^2}{b}} + \frac{a e^{ax} \sin(bx)}{b^2 + a^2}

The integral $\int e^{ax} \sin(bx) \, dx$ is simplified as

\frac{a e^{ax} \sin(bx)}{a^2 + b^2} – \frac{b e^{ax} \cos(bx)}{a^2 + b^2}

Solved Examples

Question 1

\[ \int e^x \sin(x) \, dx \]


Comparing to above formula , here a=1 and b =1, we get

\[ \int e^x \sin(x) \, dx = \frac{e^x (\sin(x) – \cos(x))}{2} + C \]

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