Since \cos(3x) is almost the derivative of \sin(3x) (the correct derivative would be 3\cos(3x)), let's multiply and divide by 3 the integral:
1/3 \int e^{\sin(3x)}3\cos(3x)\ dx
Now the integrand is of the form e^{f(x)} * f'(x), which is exactly the derivative of e^{f(x)}. So, we have nothing but
1/3 \int d/dx e^{\sin(3x)}\ dx
And since integral and derivative are one the inverse function of the other, they cancel out and the result is
1/3 e^{\sin(3x)}+c
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Another way of solving this would have been by substitution: putting t=\sin(3x), you would have dt = 3\cos(3x) dx, and the integral would have become
\int e^{t} dt/3
Factoring costants out:
1/3 \int e^t dt
But \int e^t dt=e^t+c, so 1/3 \int e^t dt = 1/3 e^t+c
Substituing back t=\sin(3x), you would obtain the same result as above.
