Question

If $I = \displaystyle \int_{-1}^{1} \frac{x^4}{1 - x^4} \cos^{-1}\left(\frac{2x}{1+x^2}\right) dx$, then $2I$ is equal to:

• A. $\pi \int_{-1}^{1} \frac{x^4}{1 - x^4} dx$
• B. $2\pi \int_{-1}^{1} \frac{x^4}{1 - x^4} dx$
• C. $\int_{-1}^{1} \frac{x^4}{1 - x^4} dx$
• D. $\pi \int_{-1}^{1} \frac{x^4}{1 + x^4} dx$
• E. $-\pi \int_{-1}^{1} \frac{x^4}{1 - x^4} dx$

Hint:

Use $f(x)+f(-x)$ trick in definite integrals with inverse trig.

Answer 1

The Correct Option is A

Concept:
• Use property: $I = \int_a^b f(x)g(x)dx$ and $I = \int_a^b f(x)g(-x)dx$

Step 1: Replace $x \to -x$
\[ I = \int_{-1}^{1} \frac{x^4}{1 - x^4} \cos^{-1}\left(\frac{-2x}{1+x^2}\right) dx \] \[ = \int_{-1}^{1} \frac{x^4}{1 - x^4} \left(\pi - \cos^{-1}\left(\frac{2x}{1+x^2}\right)\right) dx \]

Step 2: Add both expressions
\[ 2I = \pi \int_{-1}^{1} \frac{x^4}{1 - x^4} dx \] Final Conclusion:
Option (A)