Question

find intergral of e^sinx... ​

Answer 1

To Find:

ʃ [e^sinx] dx

where, y(let)=e^sinx

★Solution★

As, ʃ[e^x] dx = e^x

Here, let, u=sinx

» y=e^u

» ʃ [e^sinx] dx = e^u÷d[sinx]/dx +C

» ʃ [e^sinx] dx = [e^sinx]÷[cosx] +C

where C ∈ R