How do you integrate #inte^(sin3x) cos 3x dx#?

Question

Luke Alvarez · Accepted Answer

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.

Scarlett Allison · Answer

undefined To integrate $\int e^{\sin(3x)} \cos(3x) \, dx$, you can use integration by parts. Let $u = e^{\sin(3x)}$ and $dv = \cos(3x) \, dx$. Then, $du = 3e^{\sin(3x)} \cos(3x) \, dx$ and $v = \frac{1}{3} \sin(3x)$. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, you can integrate step by step. After the integration, you'll get your result.