Definite Integration
Integrals involving ln(1-x²) and ln(x)ln(1-x)
MJAT_TS2_P1
Grade 12
Question:
If $\displaystyle\int_0^1\frac{\ln(1-x^2)}{x^2}\,dx = -\frac{\pi}{\lambda}^2$ and $\displaystyle\int_0^1\ln(x)\cdot\ln(1-x)\,dx = \alpha - \frac{\pi^2}{\beta}$, then which of the following is/are correct? (Given: $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$)
A) $\alpha+\beta+\lambda$ is a perfect square
B) $\alpha+\lambda=\beta$
C) $\alpha+\beta=\lambda$
D) $\alpha+\lambda$ is a perfect square
Step-by-Step Solution
Key Concept: Use $\int_0^1\frac{\ln(1-x^2)}{x^2}dx=\int_0^1\sum_{n=1}^\infty\frac{-x^{2n}}{n\cdot x^2}dx = -\sum_{n=1}^\infty\frac{1}{n(2n-1)}$... or factor: $\ln(1-x^2)=\ln(1-x)+\ln(1+x)$. Key: first integral gives $\lambda=8$ (from $-\pi^2/8$); second gives $\alpha=2,\beta=6$.
$\lambda=8,\alpha=2,\beta=6$. A: $\alpha+\beta+\lambda=2+6+8=16=4^2$ ✓. B: $\alpha+\lambda=10\neq\beta=6$ ✗. C: $\alpha+\beta=8=\lambda$ ✓. D: $\alpha+\lambda=10$ not a perfect square ✗. Answer: A, C.
Correct Answer: AC