What is the integral of-e^ysin(y)cos(y) ?

The integral of-e^ysin(y)cos(y) is 1/5 e^ycos(2y)-1/10 e^ysin(2y)+C

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integral of-e^ysin(y)cos(y)

\int - e^{y} sin (y) cos (y) d y

\frac{1}{5} e^{y} cos (2 y) - \frac{1}{10} e^{y} sin (2 y) + C

Solution steps

\int - e^{y} sin (y) cos (y) d y

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

= - \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

= - \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

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

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

Therefore

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

Isolate

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

= - (- \frac{e ^{y} cos ( 2 y )}{5} + \frac{e ^{y} sin ( 2 y )}{10})

Simplify

- (- \frac{e ^{y} cos ( 2 y )}{5} + \frac{e ^{y} sin ( 2 y )}{10}) : \frac{1}{5} e^{y} cos (2 y) - \frac{1}{10} e^{y} sin (2 y)

= \frac{1}{5} e^{y} cos (2 y) - \frac{1}{10} e^{y} sin (2 y)

Add a constant to the solution

= \frac{1}{5} e^{y} cos (2 y) - \frac{1}{10} e^{y} sin (2 y) + C

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What is the integral of-e^ysin(y)cos(y) ?
The integral of-e^ysin(y)cos(y) is 1/5 e^ycos(2y)-1/10 e^ysin(2y)+C