Q. 264.5( 2 Votes )

# Evaluate the following integrals:

Answer :

Let

We know cos2θ = 1 – 2sin^{2}θ = 2cos^{2}θ – 1

Hence, in the numerator, we can write 1 – cos2x = 2sin^{2}x

In the denominator, we can write 1 + cos2x = 2cos^{2}x

Therefore, we can write the integral as

[∵ sec^{2}θ – tan^{2}θ = 1]

Recall and

∴ I = tan x – x + c

Thus,

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