Evaluate ∫_-π/4^π/4 log(sin x + cos x) dx

The correct answer is Option (1) → $\frac{\pi}{4} \log \left( \frac{1}{2} \right)$

Let $I = \int_{-\pi/4}^{\pi/4} \log(\sin x + \cos x) \, dx \dots(i)$

$\Rightarrow I = \int_{-\pi/4}^{\pi/4} \log \left\{ \sin \left( \frac{\pi}{4} - \frac{\pi}{4} - x \right) + \cos \left( \frac{\pi}{4} - \frac{\pi}{4} - x \right) \right\} dx$

$= \int_{-\pi/4}^{\pi/4} \log[\sin(-x) + \cos(-x)] \, dx$

$\Rightarrow I = \int_{-\pi/4}^{\pi/4} \log(\cos x - \sin x) \, dx \dots(ii)$

From Eqs. (i) and (ii),

$2I = \int_{-\pi/4}^{\pi/4} [\log(\cos x + \sin x) + \log(\cos x - \sin x)] dx$

$= \int_{-\pi/4}^{\pi/4} \log [(\cos x + \sin x)(\cos x - \sin x)] dx$   $[∵\log mn = \log m + \log n]$

$= \int_{-\pi/4}^{\pi/4} \log(\cos^2 x - \sin^2 x) dx$

$\Rightarrow 2I = \int_{-\pi/4}^{\pi/4} \log \cos 2x \, dx$

$\Rightarrow 2I = 2 \int_{0}^{\pi/4} \log \cos 2x \, dx \dots(iii)$

$\left[ ∵\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx, \text{if } f(-x) = f(x) \right]$

Put $2x = t \Rightarrow dx = \frac{dt}{2}$

As $x \to 0, \text{then } t \to 0$

and $x \to \frac{\pi}{4}, \text{then } t \to \frac{\pi}{2}$

$2I = \int_{0}^{\pi/2} \log \cos t \, dt \dots(iv)$

$\Rightarrow 2I = \int_{0}^{\pi/2} \log \cos \left( \frac{\pi}{2} - t \right) dt \quad \left[ ∵\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \right]$

$\Rightarrow 2I = \int_{0}^{\pi/2} \log \sin t \, dt \dots(v)$

On adding Eqs. (iv) and (v), we get

$4I = \int_{0}^{\pi/2} \log \sin t \cos t \, dt$

$\Rightarrow 4I = \int_{0}^{\pi/2} \log \frac{\sin 2t}{2} \, dt$

$\Rightarrow 4I = \int_{0}^{\pi/2} \log \sin 2x \, dx - \int_{0}^{\pi/2} \log 2 \, dx$

$\Rightarrow 4I = \int_{0}^{\pi/2} \log \sin 2x \, dx - \frac{\pi}{2} \log 2$

$\Rightarrow 4I = 2 \int_{0}^{\pi/4} \log \sin 2x \, dx - \frac{\pi}{2} \log 2$

$\left[ ∵\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \text{ if } f(2a - x) = f(x) \right]$

$\Rightarrow 4I = 2 \int_{0}^{\pi/4} \log \sin 2 \left( \frac{\pi}{4} - x \right) dx - \frac{\pi}{2} \log 2$

$= 2 \int_{0}^{\pi/4} \log \sin \left( \frac{\pi}{2} - 2x \right) dx - \frac{\pi}{2} \log 2$

$= 2 \int_{0}^{\pi/4} \log \cos 2x \, dx - \frac{\pi}{2} \log 2$

$\Rightarrow 4I = 2I - \frac{\pi}{2} \log 2 \quad [\text{using Eq. (iii)}]$

$2I = -\frac{\pi}{2} \log 2$

$I = -\frac{\pi}{4} \log 2$

$I = \frac{\pi}{4} \log \left( \frac{1}{2} \right)$