Definite IntegrationmediumFree

∫ dx/((1+x²)√(1−x²)), 0 to 1/√3 | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of the integral

\displaystyle\int_{0}^{1/\sqrt3}\dfrac{dx}{(1+x^{2})\sqrt{1-x^{2}}}

must be

\dfrac{\pi}{2\sqrt2}

\dfrac{\pi}{4\sqrt2}

correct

\dfrac{\pi}{8\sqrt2}

DNone of these

Solution

Step 1: Substitute

x=\sin\theta

dx=\cos\theta\,d\theta

. Then

\sqrt{1-x^{2}}=\cos\theta

, and the upper limit becomes

\theta=\sin^{-1}\dfrac{1}{\sqrt3}

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^{2}\theta

— equivalently, use

1+\sin^{2}\theta

written via

\csc^{2}\theta

\dfrac{d\theta}{1+\sin^{2}\theta}=\dfrac{\csc^{2}\theta\,d\theta}{\csc^{2}\theta+1}.

Step 3: Put

t=\cot\theta

dt=-\csc^{2}\theta\,d\theta

, and

\csc^{2}\theta=1+t^{2}

, so

\csc^{2}\theta+1=2+t^{2}

. The limits map as

\theta\to 0^{+}\Rightarrow t\to\infty

and

\theta=\sin^{-1}\dfrac{1}{\sqrt3}\Rightarrow t=\cot\theta=\sqrt2

(since then

\sin\theta=\tfrac{1}{\sqrt3},\cos\theta=\sqrt{\tfrac23}

). Thus

I=\int_{\infty}^{\sqrt2}\dfrac{-dt}{2+t^{2}}=\int_{\sqrt2}^{\infty}\dfrac{dt}{2+t^{2}}.

Step 4: Integrate:

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)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.