Question

$\int_{-1}^{1} \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$

• A. $\frac{\pi}{8}\log 2$
• B. $\frac{\pi}{2}\log 2$
• C. $\frac{\pi}{4}\log 2$
• D. $2\pi\log 2$

Hint:

For definite integrals involving $\sqrt{1-x^2}$, the substitution $x=\sin\theta$ or $x=\cos\theta$ is standard. Remember the important definite integral results: $\int_0^{\pi/2} \log(\sin x) dx = \int_0^{\pi/2} \log(\cos x) dx = -\frac{\pi}{2}\log 2$.

Answer 1

The Correct Option is D

Let the integral be $I$. Let's use the substitution $x = \cos\theta$.
Then $dx = -\sin\theta d\theta$.
The limits of integration change as follows:
When $x=-1$, $\cos\theta = -1 \implies \theta = \pi$.
When $x=1$, $\cos\theta = 1 \implies \theta = 0$.
The denominator is $\sqrt{1-x^2} = \sqrt{1-\cos^2\theta} = \sqrt{\sin^2\theta} = \sin\theta$ (since $\theta$ is in $[0, \pi]$, $\sin\theta \geq 0$).
Substituting into the integral:
$I = \int_{\pi}^{0} \frac{\log 2 - \log(1+\cos\theta)}{\sin\theta} (-\sin\theta d\theta)$.
The $\sin\theta$ terms cancel, and we can flip the limits of integration by removing the negative sign.
$I = \int_{0}^{\pi} (\log 2 - \log(1+\cos\theta)) d\theta$.
Using the half-angle identity $1+\cos\theta = 2\cos^2(\theta/2)$:
$I = \int_{0}^{\pi} (\log 2 - \log(2\cos^2(\theta/2))) d\theta = \int_{0}^{\pi} (\log 2 - (\log 2 + \log(\cos^2(\theta/2)))) d\theta$.
$I = \int_{0}^{\pi} - \log(\cos^2(\theta/2)) d\theta = \int_{0}^{\pi} -2\log(\cos(\theta/2)) d\theta$.
Let $u = \theta/2$. Then $d\theta = 2du$. The limits change from $0$ to $\pi/2$.
$I = \int_{0}^{\pi/2} -2\log(\cos u) (2du) = -4 \int_{0}^{\pi/2} \log(\cos u) du$.
We use the standard definite integral result $\int_{0}^{\pi/2} \log(\cos u) du = -\frac{\pi}{2}\log 2$.
$I = -4 \left(-\frac{\pi}{2}\log 2\right) = 2\pi\log 2$.