Use integration by parts to find the integral of sin(x)*exp(x)

First, we choose u=sin(x),v'=exp(x). Using differentiation and integration of standard exponential and trigonometric functions => u'=cos(x),v=exp(x). From this we use the formula for integration by parts which tells us that the integral of a product can be given by I=uv-int(vu'). Therefore I=sin(x)*exp(x)-int(exp(x)*cos(x)). Since we have another integral of a product, integration by parts must be applied again to our new integral which we can call I'=int(exp(x)*cos(x). Now, we choose u=cos(x),v'=exp(x) => u'=-sin(x),v=exp(x). Again, using the formula, we have I'= cos(x)*exp(x)-int(-sin(x)*exp(x)) I'=cos(x)*exp(x)+int(sin(x)*exp(x)). This seems to be unsolvable, since the trigonometric functions behave in a cycle under differentiation and integration, and exp(x) is unaffected. However, in this circumstance there is a trick that leads to your solution. Notice that in the equation for I' we have the integral of sin(x)*exp(x). This was what we were initially tasked with finding, and so this expression can be replaced simply with I, so I'=cos(x)*exp(x)+I. Now we have our expression for I' we can substitute it back into our equation for I, which leads to I=sin(x)*exp(x)-(cos(x)*exp(x)+I) I=sin(x)*exp(x)-cos(x)*exp(x)-I 2I=sin(x)*exp(x)-cos(x)exp(x) 2I=exp(x)[sin(x)-cos(x)] I=1/2exp(x)[sin(x)-cos(x)]+C. Ensure not to forget the constant of integration C at the end there.