Evaluate the following integrals:

Question

Philoid · Accepted 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,