Question

${\displaystyle \int \sin 4x\cos 7x dx }$

Answer 1

${\displaystyle \int \sin 4x \cos 7x dx}$
$$=\frac{1}{2}\int 2\cos 7x\sin \text{4x dx}$$
$$=\frac{1}{2}\int [\text{sin }(7x+4x)-\text{sin}(7x-4x)]dx[\therefore 2\cos A\sin B=\sin (A+B)-\sin (A-B)]$$
$$=\frac{1}{2}\int (\text{sin }\left(\text{11x}\right)-\text{sin}\left(3x\right))dx$$
$$=\frac{1}{2}[-\frac{\text{cos}\left(11x\right)}{11}+\frac{\text{cos}\left(3x\right)}{3}]+C$$
$$=-\frac{\text{cos}\left(11x\right)}{22}+\frac{\text{cos}\left(3x\right)}{6}$$