What is the integral of e^ycos(2y) ?

The integral of e^ycos(2y) is (2e^ysin(2y))/5+(e^ycos(2y))/5+C

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integral of e^ycos(2y)

\int e^{y} cos (2 y) d y

\frac{2 e ^{y} sin ( 2 y )}{5} + \frac{e ^{y} cos ( 2 y )}{5} + C

Solution steps

\int e^{y} cos (2 y) d y

Apply Integration By Parts

: e^{y} \frac{1}{2} sin (2 y) - \int e^{y} \frac{1}{2} sin (2 y) d y

= e^{y} \frac{1}{2} sin (2 y) - \int e^{y} \frac{1}{2} sin (2 y) d y

Take the constant out:

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

= e^{y} \frac{1}{2} sin (2 y) - \frac{1}{2} \cdot \int e^{y} sin (2 y) d y

Apply Integration By Parts

: - \frac{1}{2} e^{y} cos (2 y) - \int - \frac{1}{2} e^{y} cos (2 y) d y

= e^{y} \frac{1}{2} sin (2 y) - \frac{1}{2} (- \frac{1}{2} e^{y} cos (2 y) - \int - \frac{1}{2} e^{y} cos (2 y) d y)

Take the constant out:

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

= e^{y} \frac{1}{2} sin (2 y) - \frac{1}{2} (- \frac{1}{2} e^{y} cos (2 y) - (- \frac{1}{2} \cdot \int e^{y} cos (2 y) d y))

Therefore

\int e^{y} cos (2 y) d y = e^{y} \frac{1}{2} sin (2 y) - \frac{1}{2} (- \frac{1}{2} e^{y} cos (2 y) - (- \frac{1}{2} \cdot \int e^{y} cos (2 y) d y))

Isolate

\int e^{y} cos (2 y) d y

= \frac{2 e ^{y} sin ( 2 y )}{5} + \frac{e ^{y} cos ( 2 y )}{5}

Add a constant to the solution

= \frac{2 e ^{y} sin ( 2 y )}{5} + \frac{e ^{y} cos ( 2 y )}{5} + C

x \to 2 lim (\frac{2 x - 25}{x - 2})

\frac{d y}{d x}, x y = y^{2}

\int_{- \infty}^{0} \frac{1}{6 - 10 x} d x

∣ 9 x ∣, x^{2} - 10

\int q^{2} cos (7 q) d q

What is the integral of e^ycos(2y) ?
The integral of e^ycos(2y) is (2e^ysin(2y))/5+(e^ycos(2y))/5+C