Question

2. sin 3x cos 4x

Answer 1

The given function is sin3xcos4x

∫ sin3xcos4xdx (1)

Also, sinAcosB= 1 2 { sin(A+B)+sin( A−B ) }(2)

Compare (1) and (2),

A=3 and B=4

Use (2) to further solve the equations,

∫ sin3xcos4xdx = ∫ sin(3x+4x)+sin(3x−4x) dx = 1 2 ∫ { sin(7x)+sin(−x) }dx = 1 2 ∫ sin(7x)dx+ 1 2 ∫ sin(x) dx =− 1 14 cos(7x)+ 1 2 cos(x)+c

Thus, the integral of the function sin3xcos4x is − 1 14 cos(7x)+ 1 2 cos(x)+c.