Answer 1

`int_0^(pi/4) sin 4x sin 3x *dx`

= `(1)/(2) int_0^(pi/4) 2 sin 4x sin 3x *dx`

= `(1)/(2) int_0^(pi/4) [cos (4x - 3x) - cos(4x + 3x)]*dx`

= `(1)/(2) int_0^(pi/4) cos x*dx - (1)/(2) int^(pi/4)cos 7x*dx`

= `(1)/(2)[sinx]_0^(pi/4) - (1)/(2)[(sin7x)/7]_0^(pi/4)`

= `(1)/(2)[sin  pi/4 - sin 0] - (1)/(14)[sin  (7pi)/4 - sin 0]`

= `(1)/(2)[1/sqrt(2) - 0] - (1)/(14)[sin (2pi - pi/4) - 0]`

= `(1)/(2sqrt(2)) - (1)/(14)(- sin  pi/4)`

= `(1)/(2sqrt(2)) + (1)/(14sqrt(2))`

= `(7 + 1)/(14sqrt(2))`

= `(4)/(7sqrt(2))`.