What is the antiderivative of #e^(sinx)*cosx#?

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Answer 1 · 2016-03-29T19:31:41

#e^sin(x) +C#

To find

#int e^(sin x)cos x dx#

First use the substitution #u = sin x#
#-> du = cos x dx#

Now put the substitution into our expression to obtain:

#int e^u du#

Evaluating this then reversing the substitution and we get:

#e^u +C = e^sin(x) +C#