Question

sin3x*sin2x lebnitz theorem
$sin3x \times sin2x \: lebnitz \: theorem$
​

Answer 1

Answer:

sinxsin2xsin3xdx

=

1

2

∫(2sin3xsinx)sin2xdx

=

1

2

∫(cos2x-cos4x)sin2xdx

=

1

4

∫(2sin2xcos2x-2cos4xsin2x)dx

=

1

4

∫[sin4x-(sin6x-sin2x)]dx

=

1

4

∫(sin4x-sin6x+sin2x)dx

=

1

4

[-

1

4

cos4x+

1

6

cos6x-

1

2

cos2x]+c