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

Apr 5, 2015

Since $\setminus \cos \left(3 x\right)$ is almost the derivative of $\setminus \sin \left(3 x\right)$ (the correct derivative would be $3 \setminus \cos \left(3 x\right)$), let's multiply and divide by $3$ the integral:

$\frac{1}{3} \setminus \int {e}^{\setminus \sin \left(3 x\right)} 3 \setminus \cos \left(3 x\right) \setminus \mathrm{dx}$

Now the integrand is of the form ${e}^{f \left(x\right)} \cdot f ' \left(x\right)$, which is exactly the derivative of ${e}^{f \left(x\right)}$. So, we have nothing but

$\frac{1}{3} \setminus \int \frac{d}{\mathrm{dx}} {e}^{\setminus \sin \left(3 x\right)} \setminus \mathrm{dx}$

And since integral and derivative are one the inverse function of the other, they cancel out and the result is

$\frac{1}{3} {e}^{\setminus \sin \left(3 x\right)} + c$

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Another way of solving this would have been by substitution: putting $t = \setminus \sin \left(3 x\right)$, you would have $\mathrm{dt} = 3 \setminus \cos \left(3 x\right) \mathrm{dx}$, and the integral would have become

$\setminus \int {e}^{t} \frac{\mathrm{dt}}{3}$

Factoring costants out:

$\frac{1}{3} \setminus \int {e}^{t} \mathrm{dt}$

But $\setminus \int {e}^{t} \mathrm{dt} = {e}^{t} + c$, so $\frac{1}{3} \setminus \int {e}^{t} \mathrm{dt} = \frac{1}{3} {e}^{t} + c$

Substituing back $t = \setminus \sin \left(3 x\right)$, you would obtain the same result as above.