What is the integral of e^xsin(nx) ?

The integral of e^xsin(nx) is -(ne^xcos(nx))/(n^2+1)+(e^xsin(nx))/(n^2+1)+C

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integral of e^xsin(nx)

\int e^{x} sin (n x) d x

- \frac{n e ^{x} cos ( n x )}{n ^{2} + 1} + \frac{e ^{x} sin ( n x )}{n ^{2} + 1} + C

Solution steps

\int e^{x} sin (n x) d x

Apply Integration By Parts

: - e^{x} \frac{1}{n} cos (n x) - \int - e^{x} \frac{1}{n} cos (n x) d x

= - e^{x} \frac{1}{n} cos (n x) - \int - e^{x} \frac{1}{n} cos (n x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - e^{x} \frac{1}{n} cos (n x) - (- \frac{1}{n} \cdot \int e^{x} cos (n x) d x)

Apply Integration By Parts

: e^{x} \frac{1}{n} sin (n x) - \int e^{x} \frac{1}{n} sin (n x) d x

= - e^{x} \frac{1}{n} cos (n x) - (- \frac{1}{n} (e^{x} \frac{1}{n} sin (n x) - \int e^{x} \frac{1}{n} sin (n x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - e^{x} \frac{1}{n} cos (n x) - (- \frac{1}{n} (e^{x} \frac{1}{n} sin (n x) - \frac{1}{n} \cdot \int e^{x} sin (n x) d x))

Therefore

\int e^{x} sin (n x) d x = - e^{x} \frac{1}{n} cos (n x) - (- \frac{1}{n} (e^{x} \frac{1}{n} sin (n x) - \frac{1}{n} \cdot \int e^{x} sin (n x) d x))

Isolate

\int e^{x} sin (n x) d x

= - \frac{n e ^{x} cos ( n x )}{n ^{2} + 1} + \frac{e ^{x} sin ( n x )}{n ^{2} + 1}

Add a constant to the solution

= - \frac{n e ^{x} cos ( n x )}{n ^{2} + 1} + \frac{e ^{x} sin ( n x )}{n ^{2} + 1} + C

y^{^{''}} + 5 y^{^{'}} + 6 y = 12

f (x) = 2 x^{3} - 5 x^{2} + 6

\int_{- 1}^{1} x^{2} - 2 - e^{x} d x

\int_{0}^{π} sin (t) d t

y^{^{'}} = 2 x - y

What is the integral of e^xsin(nx) ?
The integral of e^xsin(nx) is -(ne^xcos(nx))/(n^2+1)+(e^xsin(nx))/(n^2+1)+C