Question

integral(sin7x sin3x )dx

Answer 1

$\displaystyle\int\sin (7x)\sin (3x)\ dx\\\\\\=\int\dfrac {1}{2}(\cos(4x)-\cos(10x))\ dx\\\\\\=\dfrac {1}{2}\int\cos(4x)\ dx-\dfrac {1}{2}\int\cos (10x)\ dx\\\\\\=\dfrac {1}{8}\int\cos (4x)\ d(4x)-\dfrac {1}{20}\int\cos (10x)\ d(10x)\\\\\\=\dfrac {\sin(4x)}{8}-\dfrac {\sin (10x)}{20}\\\\\\=\mathbf {\dfrac {5\sin (4x)-2\sin (10x)}{40}}$

Answer 2

guys this is the ans for this question hope you will understand