Evaluate
We can write above integral as:
--(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,
Cosidering
We know,
2 sinA.cosB = sin(A+B) + sin(A-B)
Now, considering A as 3x and B as x we get,
2 sin3x.cosx = sin(4x) + sin(2x)
--(2)
Again, Cosidering
We know,
2 sinA.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)
--(3)
∴ integral becomes,
[From (2) and (3)]