Question

Evaluate ∫ cosx cos2x cos3x dx

Answer 1

∫ cosx cos2x cos3x dx

We can write above integral as:

= \(\frac{1}{2}\)∫(2cosx cos2x) cos3x dx ...(1)

We know that, 

2 cosA.cosB = cos(A+B) + cos(A-B) 

Now, 

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

= 2 cosx.cos2x 

= cos(x+2x) + cos(x-2x) 

= 2 cosx.cos2x 

= cos(3x) + cos(-x) 

= 2 cosx.cos2x 

= cos(3x) + cos(x) 

[∵ cos(-x) = cos(x)] 

∴ integral (1) becomes,

We know, 

2 cosA.cosB = cos(A+B) + cos(A-B) 

Now, 

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

2 cosx.cos3x = cos(4x) + cos(-2x) 

2 cosx.cos3x = cos(4x) + cos(2x) 

[∵ cos(-x) = cos(x)]

∴ ∫2(cosx.cos3x)dx 

= ∫(cos4x + cos2x) dx...(2)

Considering, 

∫ 2cos²3x 

We know, 

cos2A = 2cos²A – 1 

2cos²A = 1 + cos2A 

Now, 

Considering A as 3x we get, 

∫ 2cos²3x = ∫ 1 + cos2(3x) 

= ∫ 1 + cos(6x)