What is the integral of e^{xy}(ycos(x)-sin(x)) ?

The integral of e^{xy}(ycos(x)-sin(x)) is e^{yx} 1/(1+y^2)cos(x)+e^{yx}(y^2)/(1+y^2)cos(x)+C

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integral of e^{xy}(ycos(x)-sin(x))

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

e^{y x} \frac{1}{1 + y ^{2}} cos (x) + e^{y x} \frac{y ^{2}}{1 + y ^{2}} cos (x) + C

Solution steps

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

Expand

(- sin (x) + cos (x) y) e^{y x} : - e^{y x} sin (x) + y e^{y x} cos (x)

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

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

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

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

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

Simplify

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

= e^{y x} \frac{1}{1 + y ^{2}} cos (x) + e^{y x} \frac{y ^{2}}{1 + y ^{2}} cos (x)

Add a constant to the solution

= e^{y x} \frac{1}{1 + y ^{2}} cos (x) + e^{y x} \frac{y ^{2}}{1 + y ^{2}} cos (x) + C

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What is the integral of e^{xy}(ycos(x)-sin(x)) ?
The integral of e^{xy}(ycos(x)-sin(x)) is e^{yx} 1/(1+y^2)cos(x)+e^{yx}(y^2)/(1+y^2)cos(x)+C