Evaluate the following Integrals:


Let


As we have the trigonometric identity, to evaluate this integral we use x = cos 2θ


dx = –2sin(2θ)dθ (Differentiating both sides)


When x = 0, cos 2θ = 0 2θ = θ =


When x = 1, cos 2θ = 1 2θ = 0 θ = 0


So, the new limits are and 0.


Substituting this in the original integral,





We have and sin 2θ = 2 sin θ cos θ




But,











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