Question

Evaluate ∫ sinx sin2x sin3x dx

Answer 1

∫ sinx sin2x sin3x dx

We can write above integral as:

= \(\frac{1}{2}\)∫(2 sinx sin2x) sin3x dx ...(1)

We know that, 

2 sinA.sinB = cos(A-B) – cos(A+B) 

Now, 

Considering A as x and B as 2x we get, 

= 2 sinx.sin2x 

= cos(x-2x) – cos(x+2x) 

= 2 sinx.sin2x 

= cos(-x) – cos(3x) 

= 2 sinx.sin2x 

= cos(x) – cos(3x) 

[∵ cos(-x) = cos(x)] 

∴ integral (1) becomes,

Considering, 

∫2(cosx.sin3x)dx

We know, 

2sinA.cosB = sin(A+B) + sin(A-B) 

Now, 

Considering A as 3x and B as x we get, 

2sin3x.cosx = sin(4x) + sin(2x)

∴ ∫2(cosx.sin3x)dx = ∫sin4x+ sin2x dx ...(2)

Again, Considering

∫2(cos3x.sin3x)dx

We know, 

2sinA.cosB = sin(A+B) + sin(A-B) 

Now, 

Considering A as 3x and B as 3x we get, 

2 sin3x.cos3x = sin(6x) + sin(0) 

= sin(6x)