What is the integral of cos(ax)e^{bx} ?

The integral of cos(ax)e^{bx} is (ae^{bx}sin(ax))/(a^2+b^2)+(be^{bx}cos(ax))/(a^2+b^2)+C

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integral of cos(ax)e^{bx}

\int cos (a x) e^{b x} d x

\frac{a e ^{b x} sin ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x} cos ( a x )}{a ^{2} + b ^{2}} + C

Solution steps

\int cos (a x) e^{b x} d x

Apply Integration By Parts

= e^{b x} \frac{1}{a} sin (a x) - \int e^{b x} b \frac{1}{a} sin (a x) d x

Take the constant out:

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

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

Simplify

b \frac{1}{a} : \frac{b}{a}

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

Apply Integration By Parts

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

Take the constant out:

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

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

Simplify

- b \frac{1}{a} : - \frac{b}{a}

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

Therefore

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

Isolate

\int e^{b x} cos (a x) d x

= \frac{a e ^{b x} sin ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x} cos ( a x )}{a ^{2} + b ^{2}}

Add a constant to the solution

= \frac{a e ^{b x} sin ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x} cos ( a x )}{a ^{2} + b ^{2}} + C

What is the integral of cos(ax)e^{bx} ?
The integral of cos(ax)e^{bx} is (ae^{bx}sin(ax))/(a^2+b^2)+(be^{bx}cos(ax))/(a^2+b^2)+C