Evaluate $\int sin3x$ $cos4x$ $dx$

The correct answer is $ \dfrac{cosx}{2} - \dfrac{cos7x}{14} $ + $C$

$\int sin3x$ $cos4x$ $dx$

Here we use the formula $sin A cos B$ = $\dfrac{1}{2} \left(sin(A+B) + sin (A-B) \right)$

In our case, $A = 3x$ and $B = 4x$

∴ $\int sin3x$ $cos4x$ $dx$

= $\dfrac{1}{2} \int sin7x$ $+ sin(-x)$ $dx$

= $\dfrac{1}{2} \int sin7x$ $dx$ $- \int sin(x)$ $dx$

= $\dfrac{1}{2}$ $\left[ \dfrac{-cos7x}{7} + cos x \right]$ + $C$

= $ \dfrac{cosx}{2} - \dfrac{cos7x}{14} $ + $C$