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)]







1