Let I = \(\int\limits_{0}^{1/2}\cfrac{1}{(1+\text x^2)\sqrt{1-\text x^2}}d\text x \)
Put x = sin θ
⇒ dx = cos θ dθ (Differentiating both sides)
Also, \(\sqrt{1-\text x^2}=\sqrt{1-sin^2\theta}\) = cos \(\theta\)
When x = 0, sin θ = 0 ⇒ θ = 0
When x = \(\cfrac12\), sin θ = \(\cfrac12\) ⇒ θ = \(\cfrac{\pi}6\)
So, the new limits are 0 and \(\cfrac{\pi}6\)
Substituting this in the original integral,
Dividing numerator and denominator with cos2θ, we have
[∵ sec2θ = 1 + tan2θ]
Put tan θ = t
⇒ sec2θ dθ = dt (Differentiating both sides)
When θ = 0, t = tan 0 = 0
When θ = \(\cfrac{\pi}6\), t = tan \(\cfrac{\pi}6\) = \(\cfrac{1}{\sqrt3}\)
So, the new limits are 0 and \(\cfrac{1}{\sqrt3}\)
Substituting this in the original integral,
Recall