What is the integral of e^tcos(nt) ?

The integral of e^tcos(nt) is (ne^tsin(nt))/(n^2+1)+(e^tcos(nt))/(n^2+1)+C

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integral of e^tcos(nt)

\int e^{t} cos (n t) d t

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

Solution steps

\int e^{t} cos (n t) d t

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{t} cos (n t) d t

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

Add a constant to the solution

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

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What is the integral of e^tcos(nt) ?
The integral of e^tcos(nt) is (ne^tsin(nt))/(n^2+1)+(e^tcos(nt))/(n^2+1)+C