Evaluate the following integrals:
Let 
We know cos2θ = 1 – 2sin2θ
Hence, in the denominator, we can write 1 – cos2x = 2sin2x
In the numerator, we have sin2x = 2sinxcosx
Therefore, we can write the integral as
Recall 
Thus, 
37
Evaluate the following integrals:
Let
We know cos2θ = 1 – 2sin2θ
Hence, in the denominator, we can write 1 – cos2x = 2sin2x
In the numerator, we have sin2x = 2sinxcosx
Therefore, we can write the integral as
Recall
Thus,