Q. 215.0( 1 Vote )

Evaluate

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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