Definite Integration
Integrals with arctan and log-sin
MJAT_TS4_P2
Grade 12
Question:
Which of the following statements are TRUE?
A) $\displaystyle\int_0^1\frac{\tan^{-1}x}{x\sqrt{1-x^2}}\,dx = \frac{\pi}{2}\ln(1+\sqrt{2})$
B) $\displaystyle\int_0^1\frac{\tan^{-1}x}{x\sqrt{1-x^2}}\,dx = \frac{\pi}{2}\ln(\sqrt{2}-1)$
C) $\displaystyle\int_0^{\pi/2}\log(\sin x)\,dx = -\frac{\pi}{2}\ln 2$
D) $\displaystyle\int_0^1\frac{\tan^{-1}x}{x\sqrt{1-x^2}}\,dx = \bigl(\log_2(\sqrt{2}-1)\bigr)\displaystyle\int_0^{\pi/2}\log(\sin x)\,dx$
Step-by-Step Solution
Key Concept: C: $\int_0^{\pi/2}\ln(\sin x)dx=-\frac{\pi}{2}\ln 2$ (classic result). A: $\int_0^1\frac{\tan^{-1}x}{x\sqrt{1-x^2}}dx$. Let $x=\sin\theta$: $=\int_0^{\pi/2}\frac{\tan^{-1}(\sin\theta)}{\sin\theta}d\theta$. Using differentiation under integral sign: $=\frac{\pi}{2}\ln(1+\sqrt{2})$ (A ✓).
A ✓, B ✗, C ✓, D ✓. Answer: A, C, D.
Correct Answer: ACD