Q. 215.0( 1 Vote )

Evaluate<span lan

Answer :


We can write above integral as:


--(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,






Cosidering


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)


Cosidering ∫ 2cos23x


We know,


cos2A = 2cos2A – 1


2cos2A = 1 + cos2A


Now, considering A as 3x we get,


∫ 2cos23x = ∫ 1 + cos2(3x) = ∫ 1 + cos(6x)


--(3)


integral becomes,



[From (2) and (3)]






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