Evaluate the following Integrals:
Let
As we have the trigonometric identity
to evaluate this integral we use x = acos 2θ
⇒ dx = –2a sin(2θ) dθ (Differentiating both sides)
When x = –a, acos 2θ = –a ⇒ cos 2θ = –1
When x = a, acos 2θ = a ⇒ cos 2θ = 1
⇒ 2θ = 0 ⇒ θ = 0
So, the new limits are and 0.
Also,
Substituting this in the original integral,
But, we have 2 sin2θ = 1 – cos 2θ