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

The integral of e^{ax}cos(b)x is (e^{ax}(-1+ax))/(a^2sec(b))+C

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integral of e^{ax}cos(b)x

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

\frac{e ^{a x} ( - 1 + a x )}{a ^{2} sec ( b )} + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

\int e^{a x} \frac{1}{a} d x = e^{a x} \frac{1}{a ^{2}}

= cos (b) (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}})

Simplify

cos (b) (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}}) : \frac{e ^{a x} ( - 1 + a x )}{a ^{2} sec ( b )}

= \frac{e ^{a x} ( - 1 + a x )}{a ^{2} sec ( b )}

Add a constant to the solution

= \frac{e ^{a x} ( - 1 + a x )}{a ^{2} sec ( b )} + C

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What is the integral of e^{ax}cos(b)x ?
The integral of e^{ax}cos(b)x is (e^{ax}(-1+ax))/(a^2sec(b))+C