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,