Integration by parts; ∫e^x sin(x) dx

The first step in solving this problem is to identify that it is an integral that needs to be solved by parts. When you see an integral, check if it is in the form uv, where u and v are both functions of x (although in special cases we may choose u=1).
In this case we can see that the integral is the product of e^x and sin(x) so we shall proceed to solve it by parts;
1) Choose u and v’2)Work out u’ and v3) Insert into the parts formula 4) Solve the resulting (simpler) integral5) Put it all together
1) This is the most important step as a mistake here can make the integral seem impossible. In general you want to choose u so that it becomes simpler when you differentiate it, as this will result in a solvable integral.
This is not always obvious however, indeed in this case, neither e^x or sin(x) appear to become simpler when you differentiate. In this case it doesn’t matter but in general a good rule to follow is LATE;LogarithmsAlgebra (polynomials like x² )Trigonometry Exponentials
In this case we have no logs or algebra so we shall choose sin(x) as our u.
2) These are both standard results; u’=cos(x)v’= e^x so v=e^x3) The formula is;∫ uv’= uv - ∫u’vSubstituting in our functions;∫ e^xsin(x) dx= e^xsin(x) - ∫e^xcos(x) dx**This looks just as bad as our original integral and many students will despair at this stage, you must be sure of your technique and persevere.... ∫e^xcos(x) dxWe are going to have to do integration by parts all over again!!
u=cos(x) u’=-sin(x)v’=e^x v = e^x ∫e^xcos(x) dx = e^xcos(x) - ∫-e^xsin(x) dxAgain another integral by parts!!! But wait, this is our original integral. Let’s put it all together;
∫ e^xsin(x) dx= e^xsin(x) - (e^xcos(x) - ∫-e^xsin(x) dx) = e^x(sin(x)-cos(x)) - ∫-e^xsin(x) dx
adding ∫-e^xsin(x) dx to both sides;
2 ∫-e^xsin(x) dx = e^x(sin(x)-cos(x))
and finally dividing by 2;
∫-e^xsin(x) dx = e^x(sin(x)-cos(x))/2