Question

If \(f(x)=\int \frac{x^{2}}{(1-x^{2})(1+\sqrt{1-x^{2}})}dx\) and \(f(0)=2\), then \(f\left(\frac{1}{2}\right)=\) ...

• A. \(\frac{1}{2}\log 3-\frac{\pi }{3}+2\)
• B. \(\frac{1}{2}\log 3+\frac{\pi }{3}+2\)
• C. \(\frac{1}{2}\log 3-\frac{\pi }{6}+2\)
• D. \(\frac{1}{2}\log 3+\frac{\pi }{6}+2\)

Hint:

Rationalise the denominator to get 1/(1-x^2) minus 1/sqrt(1-x^2), integrate, then use f(0) = 2.

Answer 1

The Correct Option is C

Step 1: Substitution route
Let $x=\sin \theta$. The integrand becomes $\dfrac{\sin ^{2}\theta }{\cos ^{2}\theta (1+\cos \theta )}\cos \theta \,d\theta $.

Step 2: Result
This reduces to $\sec \theta -1$-type terms, giving $\log |\sec \theta +\tan \theta |-\theta$, the same as $\dfrac{1}{2}\log \dfrac{1+x}{1-x}-\sin ^{-1}x$.

Step 3: Evaluate
At $x=\dfrac{1}{2}$: $\dfrac{1}{2}\log 3-\dfrac{\pi }{6}$. Adding the constant $2$ gives the answer.

Final Answer:
Adding the constant 2 to $\dfrac{1}{2}\log 3-\dfrac{\pi }{6}$ gives the result. \[ \boxed{\text{(C) }\dfrac{1}{2}\log 3-\dfrac{\pi }{6}+2} \]