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,


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