If $\int_0^{\pi/4} x\cos\left(\frac{b}{1+x^2}\right)dx = k\int_0^{b} \frac{1+u^2}{1-u^2}du$, then the value of $k$ is equal to
Step-by-Step Solution
Key Concept: For odd functions integrated over symmetric intervals $[-a, a]$, special properties can simplify the computation using substitution and symmetry.
Let $I = \int_{-\pi/3}^{\pi/3} \left(\frac{x}{1+x^2}\right) \cos^{-1}\left(\frac{2x}{1+x^2}\right) dx$. The integrand contains $\frac{x}{1+x^2}$ which is an odd function multiplied by $\cos^{-1}$ term. Since $\frac{x}{1+x^2}$ is odd, we use the property that $\int_{-a}^a \text{(odd function)} \, dx = \frac{2}{\pi} \int_0^\pi f dx = \pi$.
Correct Answer: π