# How do I evaluate the integral intsin4x cos2x dx?

Probably, there are various ways to find the antiderivative(s) of $f \left(x\right) = \sin \left(4 x\right) \cos \left(2 x\right)$. I suggest integration by parts.
In fact, when trying and integrate the product of functions that exhibit cyclic derivatives, integration by parts is often a very powerful technique.

Recall that integration by parts comes from the chain rule and states the following: given $h \left(x\right)$ and $k \left(x\right)$ continuously differentiable functions, then

$\int h \left(x\right) k ' \left(x\right) \mathrm{dx} = h \left(x\right) k \left(x\right) - \int h ' \left(x\right) k \left(x\right) \mathrm{dx}$

In our case, we choose $h \left(x\right) = \sin \left(4 x\right)$ and $k ' \left(x\right) = \cos \left(2 x\right)$:

$\int f \left(x\right) \mathrm{dx} = \sin \left(4 x\right) \sin \frac{2 x}{2} - \int 4 \cos \left(4 x\right) \sin \frac{2 x}{2} \mathrm{dx} = \frac{1}{2} \sin \left(4 x\right) \sin \left(2 x\right) - 2 \int \cos \left(4 x\right) \sin \left(2 x\right) \mathrm{dx}$

and by parts again on $2 \int \cos \left(4 x\right) \sin \left(2 x\right) \mathrm{dx}$, choosing $h \left(x\right) = \cos \left(4 x\right)$ and $k ' \left(x\right) = 2 \sin \left(2 x\right)$:

$\int \cos \left(4 x\right) 2 \sin \left(2 x\right) \mathrm{dx} = \cos \left(4 x\right) \left[- \cos \left(2 x\right)\right] - \int \left[- 4 \sin \left(4 x\right)\right] \left[- \cos \left(2 x\right)\right] \mathrm{dx} = - \cos \left(4 x\right) \cos \left(2 x\right) - 4 \int \sin \left(4 x\right) \cos \left(2 x\right) \mathrm{dx} = - \cos \left(4 x\right) \cos \left(2 x\right) - 4 \int f \left(x\right) \mathrm{dx}$

Finally, substituting the expression obtained for $2 \int \cos \left(4 x\right) \sin \left(2 x\right) \mathrm{dx}$ into the expression calculated for $\int f \left(x\right) \mathrm{dx}$, we get:

$\int f \left(x\right) \mathrm{dx} = \frac{1}{2} \sin \left(4 x\right) \sin \left(2 x\right) + \cos \left(4 x\right) \cos \left(2 x\right) + 4 \cdot \int f \left(x\right) \mathrm{dx}$

$3 \int f \left(x\right) \mathrm{dx} = - \frac{1}{2} \sin \left(4 x\right) \sin \left(2 x\right) - \cos \left(4 x\right) \cos \left(2 x\right) + t i l \mathrm{de} \left\{C\right\}$

$\int f \left(x\right) \mathrm{dx} = - \frac{1}{6} \sin \left(4 x\right) \sin \left(2 x\right) - \frac{1}{3} \cos \left(4 x\right) \cos \left(2 x\right) + C$

This result can be rewritten in many ways using trigonometric identities.