Q. 264.5( 2 Votes )

Evaluate the following integrals:


Answer :

Let


We know cos2θ = 1 – 2sin2θ = 2cos2θ – 1


Hence, in the numerator, we can write 1 – cos2x = 2sin2x


In the denominator, we can write 1 + cos2x = 2cos2x


Therefore, we can write the integral as





[ sec2θ – tan2θ = 1]



Recall and


I = tan x – x + c


Thus,


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