Evaluate the following integrals:

Let

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

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

In the denominator, we can write 4x = 2 × 2x

⇒ 1 + cos4x = 1 + cos(2×2x)

⇒ 1 + cos4x = 2cos²2x

Therefore, we can write the integral as

Recall

Thus,