Question

integral por partes de sen4x. sen3x dx......

Answer 1

se debe usar identidades trigonométicas

$\cos(x-y)-\cos(x+y)=2\sin x\sin y\\ \\ \text{Por ende: }\\ \sin4x\sin3x = \dfrac{1}{2}\left(\cos x -\cos 7x\right)\\ \\ \text{As\'i:}\\ \\ \displaystyle \int\sin4x\sin3x \,dx=\dfrac{1}{2}\int \cos x-\cos 7x \, dx\\ \\ \int\sin4x\sin3x \,dx=\dfrac{1}{2}\left(\sin x - \dfrac{1}{7}\sin7x\right)+C\\ \\ \\ \boxed{\int\sin4x\sin3x \,dx=\dfrac{1}{2}\sin x - \dfrac{1}{14}\sin7x+C}$