Question

The value of the integral
\(\displaystyle\int_{-1}^{1}\log_e\left(\sqrt{1-x}+\sqrt{1+x}\right)dx\)
is equal to:

• A. \(\frac{1}{2}\log_e 2+\frac{\pi}{4}-\frac{3}{2}\)
• B. \(2\log_e 2+\frac{\pi}{4}-1\)
• C. \(\log_e 2+\frac{\pi}{2}-1\)
• D. \(2\log_e 2+\frac{\pi}{2}-\frac{1}{2}\)

Hint:

Square the argument of the logarithm to get \(2+2\sqrt{1-x^2}\), use evenness, then substitute \(x=\sin\theta\) and integrate by parts.

Answer 1

The Correct Option is C

Since \(\left(\sqrt{1-x}+\sqrt{1+x}\right)^2=2+2\sqrt{1-x^2}\), the integrand is \(\frac12\log_e\left(2+2\sqrt{1-x^2}\right)\), an even function.
So \(I=2\cdot\frac12\int_0^1\log_e\left(2+2\sqrt{1-x^2}\right)dx=\int_0^1\log_e\left(2+2\sqrt{1-x^2}\right)dx\).
Put \(x=\sin\theta\): \(I=\int_0^{\pi/2}\log_e(2+2\cos\theta)\cos\theta\,d\theta=\int_0^{\pi/2}\left[2\log_e 2+2\log_e\cos\frac{\theta}{2}\right]\cos\theta\,d\theta\).
First part: \(2\log_e 2\int_0^{\pi/2}\cos\theta\,d\theta=2\log_e 2\).
Second part, by parts with \(u=\log_e\cos\frac\theta2,\ dv=\cos\theta\,d\theta\): \(\int_0^{\pi/2}\log_e\cos\frac\theta2\cos\theta\,d\theta=\left[\sin\theta\log_e\cos\frac\theta2\right]_0^{\pi/2}+\frac12\int_0^{\pi/2}\sin\theta\tan\frac\theta2\,d\theta\).
The bracket equals \(\log_e\frac{1}{\sqrt2}=-\frac12\log_e2\). Also \(\sin\theta\tan\frac\theta2=2\sin^2\frac\theta2=1-\cos\theta\), so the last integral is \(\frac12\left(\frac\pi2-1\right)\).
So the second part is \(2\left[-\frac12\log_e 2+\frac\pi4-\frac12\right]=-\log_e2+\frac\pi2-1\).
\(I=2\log_e2-\log_e2+\frac\pi2-1=\log_e2+\frac\pi2-1\).
Answer: (3) \(\log_e 2+\frac{\pi}{2}-1\)