Integrating exponential times trigonometric functions

This lesson integrates a product of an exponential function and a trigonometric function. The key idea is that applying integration by parts twice brings the original integral back, so you can treat it as an unknown, move it to the other side, and solve for it algebraically.

Example 1: $\int e^{2x}\sin 3x\,dx$

Call the integral $I$. Take the first function as $e^{2x}$ and the second as $\sin 3x$, then apply integration by parts.

$I = e^{2x}\left(-\tfrac{\cos 3x}{3}\right) - \int 2e^{2x}\left(-\tfrac{\cos 3x}{3}\right)dx$

$I = -\tfrac{1}{3}e^{2x}\cos 3x + \tfrac{2}{3}\int e^{2x}\cos 3x\,dx$

Second integration by parts

Apply parts again to $\int e^{2x}\cos 3x\,dx$, with first function $e^{2x}$ and second $\cos 3x$.

$\int e^{2x}\cos 3x\,dx = e^{2x}\cdot\tfrac{\sin 3x}{3} - \int 2e^{2x}\cdot\tfrac{\sin 3x}{3}\,dx$

Substituting back:

$I = -\tfrac{1}{3}e^{2x}\cos 3x + \tfrac{2}{9}e^{2x}\sin 3x - \tfrac{4}{9}\int e^{2x}\sin 3x\,dx$

The last integral is $I$ itself.

Solving for $I$

$I = -\tfrac{1}{3}e^{2x}\cos 3x + \tfrac{2}{9}e^{2x}\sin 3x - \tfrac{4}{9}I$

Transpose $-\tfrac{4}{9}I$ to the left:

$I + \tfrac{4}{9}I = \tfrac{13}{9}I = \tfrac{1}{9}e^{2x}\left(2\sin 3x - 3\cos 3x\right)$

$I = \tfrac{1}{13}e^{2x}\left(2\sin 3x - 3\cos 3x\right) + C$

Example 2: $\int e^{x}\cos x\,dx$

Call it $I$. Take the first function as $e^{x}$ and the second as $\cos x$.

$I = e^{x}\sin x - \int e^{x}\sin x\,dx$

Apply parts again to $\int e^{x}\sin x\,dx$:

$\int e^{x}\sin x\,dx = -e^{x}\cos x + \int e^{x}\cos x\,dx$

Substituting back, the last integral is again $I$:

$I = e^{x}\sin x + e^{x}\cos x - I$

$2I = e^{x}\left(\sin x + \cos x\right)$

$I = \tfrac{1}{2}e^{x}\left(\sin x + \cos x\right) + C$

Key takeaways

• For $\int e^{ax}\sin bx\,dx$ and $\int e^{ax}\cos bx\,dx$, integrate by parts twice and the original integral returns.
• Once it returns, move it to the other side and divide to solve for $I$ directly.
• $\int e^{2x}\sin 3x\,dx = \tfrac{1}{13}e^{2x}(2\sin 3x - 3\cos 3x) + C$ and $\int e^{x}\cos x\,dx = \tfrac{1}{2}e^{x}(\sin x + \cos x) + C$.