Question

what the ans of intigration( e^y cos x) dx

Answer 1

Answer:

dydx=ycos(xy)+eysinxeycosx−xcos(xy)
Explanation:

Given: eycosx=1+sin(xy)

We shall use the Product Rule :

If u(x) and v(x) are two functions of x, then ddx(u⋅v)=u⋅dvdx+dudx⋅v

Differentiation: ey⋅(−sinx)+eycosx⋅dydx=0+cos(xy)(xdydx+y)

−eysinx+eycosx⋅dydx=xcos(xy)⋅dydx+ycos(xy)

eycosx⋅dydx−xcos(xy)⋅dydx=ycos(xy)+eysinx

Factoring ouy dydx on the left hand side we have

[eycosx−xcos(xy)]dydx=ycos(xy)+eysinx

Finally, dydx=ycos(xy)+eysinxeycosx−xcos(xy)