What is the integral of e^{5x}cos^2(x) ?

The integral of e^{5x}cos^2(x) is 1/290 (29e^{5x}+10e^{5x}sin(2x)+25e^{5x}cos(2x))+C

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integral of e^{5x}cos^2(x)

\int e^{5 x} cos^{2} (x) d x

\frac{1}{290} (29 e^{5 x} + 10 e^{5 x} sin (2 x) + 25 e^{5 x} cos (2 x)) + C

Solution steps

\int e^{5 x} cos^{2} (x) d x

Rewrite using trig identities

= \int \frac{1}{2} e^{5 x} (1 + cos (2 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{5 x} (1 + cos (2 x)) d x

Expand

e^{5 x} (1 + cos (2 x)) : e^{5 x} + e^{5 x} cos (2 x)

= \frac{1}{2} \cdot \int e^{5 x} + e^{5 x} cos (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int e^{5 x} d x + \int e^{5 x} cos (2 x) d x)

\int e^{5 x} d x = \frac{1}{5} e^{5 x}

\int e^{5 x} cos (2 x) d x = \frac{2 e ^{5 x} sin ( 2 x )}{29} + \frac{5 e ^{5 x} cos ( 2 x )}{29}

= \frac{1}{2} (\frac{1}{5} e^{5 x} + \frac{2 e ^{5 x} sin ( 2 x )}{29} + \frac{5 e ^{5 x} cos ( 2 x )}{29})

Simplify

\frac{1}{2} (\frac{1}{5} e^{5 x} + \frac{2 e ^{5 x} sin ( 2 x )}{29} + \frac{5 e ^{5 x} cos ( 2 x )}{29}) : \frac{1}{290} (29 e^{5 x} + 10 e^{5 x} sin (2 x) + 25 e^{5 x} cos (2 x))

= \frac{1}{290} (29 e^{5 x} + 10 e^{5 x} sin (2 x) + 25 e^{5 x} cos (2 x))

Add a constant to the solution

= \frac{1}{290} (29 e^{5 x} + 10 e^{5 x} sin (2 x) + 25 e^{5 x} cos (2 x)) + C

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What is the integral of e^{5x}cos^2(x) ?
The integral of e^{5x}cos^2(x) is 1/290 (29e^{5x}+10e^{5x}sin(2x)+25e^{5x}cos(2x))+C