Question

sinx. sin 2x. sin 3x differentiation with respect to x​

Answer 1

Answer:

Let I=∫sinx.sin2x.sin3xdx

=

2

1

∫2sinx.sin2x.sin3xdx

=

2

1

∫sinx.(2sin2x.sin3x)dx

We know that [2sinA.sinB=cos(A−B)−cos(A+B)]

So, =

2

1

∫sinx.(cosx−cos5x)dx

=

2×2

1

∫2sinx.cosxdx−

2×2

1

∫2sinx.cos5xdx

=

4

1

∫sin2xdx−

4

1

∫(sin6x−sin4x)dx

= −

8

cos2x

+

24

cos6x

−

16

cos4x

+c