# Evaluate the following integrals:

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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