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∫ dx/((1+x²)√(1−x²)), 0 to 1/√3 | JEE

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Question
The value of the integral ∫01/3dx(1+x2)1−x2\displaystyle\int_{0}^{1/\sqrt3}\dfrac{dx}{(1+x^{2})\sqrt{1-x^{2}}} must be
Aπ22\dfrac{\pi}{2\sqrt2}
Bπ42\dfrac{\pi}{4\sqrt2}correct
Cπ82\dfrac{\pi}{8\sqrt2}
DNone of these
Solution
Step 1: Substitute x=sin⁡θx=\sin\theta, dx=cos⁡θ dθdx=\cos\theta\,d\theta. Then 1−x2=cos⁡θ\sqrt{1-x^{2}}=\cos\theta, and the upper limit becomes θ=sin⁡−113\theta=\sin^{-1}\dfrac{1}{\sqrt3}:
I=∫0sin⁡−1(1/3)cos⁡θ dθ(1+sin⁡2θ)cos⁡θ=∫0sin⁡−1(1/3)dθ1+sin⁡2θ.I=\int_{0}^{\sin^{-1}(1/\sqrt3)}\dfrac{\cos\theta\,d\theta}{(1+\sin^{2}\theta)\cos\theta}=\int_{0}^{\sin^{-1}(1/\sqrt3)}\dfrac{d\theta}{1+\sin^{2}\theta}.
 Step 2: Convert to a cotangent form. Divide numerator and denominator by sin⁡2θ\sin^{2}\theta — equivalently, use 1+sin⁡2θ1+\sin^{2}\theta written via csc⁡2θ\csc^{2}\theta:
dθ1+sin⁡2θ=csc⁡2θ dθcsc⁡2θ+1.\dfrac{d\theta}{1+\sin^{2}\theta}=\dfrac{\csc^{2}\theta\,d\theta}{\csc^{2}\theta+1}.
 Step 3: Put t=cot⁡θt=\cot\theta, dt=−csc⁡2θ dθdt=-\csc^{2}\theta\,d\theta, and csc⁡2θ=1+t2\csc^{2}\theta=1+t^{2}, so csc⁡2θ+1=2+t2\csc^{2}\theta+1=2+t^{2}. The limits map as θ→0+⇒t→∞\theta\to 0^{+}\Rightarrow t\to\infty and θ=sin⁡−113⇒t=cot⁡θ=2\theta=\sin^{-1}\dfrac{1}{\sqrt3}\Rightarrow t=\cot\theta=\sqrt2 (since then sin⁡θ=13,cos⁡θ=23\sin\theta=\tfrac{1}{\sqrt3},\cos\theta=\sqrt{\tfrac23}). Thus
I=∫∞2−dt2+t2=∫2∞dt2+t2.I=\int_{\infty}^{\sqrt2}\dfrac{-dt}{2+t^{2}}=\int_{\sqrt2}^{\infty}\dfrac{dt}{2+t^{2}}.
 Step 4: Integrate:
I=12[tan⁡−1t2]2∞=12(π2−π4)=12⋅π4=π42.I=\dfrac{1}{\sqrt2}\Big[\tan^{-1}\dfrac{t}{\sqrt2}\Big]_{\sqrt2}^{\infty}=\dfrac{1}{\sqrt2}\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt2}\cdot\dfrac{\pi}{4}=\dfrac{\pi}{4\sqrt2}.
 Correct answer: (2)
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