$$\begin{aligned} \text{Let}\quad I & =\int_{-\pi / 4}^{\pi / 4} \log (\sin x+\cos x) d x \quad\text{.... (i)}\\ I & =\int_{-\pi / 4}^{\pi / 4} \log \left\{\sin \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}-\frac{\pi}{4}-x\right)\right\} d x \\ & =\int_{-\pi / 4}^{\pi / 4} \log \{\sin (-x)+\cos (-x)\} d x \\ \text{and}\quad I & =\int_{-\pi / 4}^{\pi /} \log (\cos x-\sin x) d x\quad\text{.... (ii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned} & 2 I=\int_{-\pi / 4}^{\pi 4} \log \cos 2 x d x \\ & 2 I=\int_0^{\pi / 4} \log \cos 2 x d x\quad\text{.... (iii)} \end{aligned}$$

$\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x)\right.$, if $\left.f(-x)=f(x)\right]$

Put $$2 x=t \Rightarrow d x=\frac{d t}{2}$$

As $x \rightarrow 0$, then $t \rightarrow 0$

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

$$\begin{aligned} & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos t d t \quad\text{.... (iv)}\\ & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos \left(\frac{\pi}{2}-t\right) d t \quad \left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]\\ \Rightarrow\quad & 2 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t d x\quad\text{.... (v)} \end{aligned}$$

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

$$\begin{aligned} & 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin t \cos t d t \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \frac{\sin 2 t}{2} d t \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin 2 x d x-\frac{1}{2} \int_0^{\pi / 2} \log 2 d x \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \sin \left(\frac{\pi}{2}-2 x\right) d x-\log 2 \cdot \frac{\pi}{4} \\ & \Rightarrow \quad 4 I=\frac{1}{2} \int_0^{\pi / 2} \log \cos 2 x d x-\frac{\pi}{4} \log 2 \end{aligned}$$

$$\begin{array}{llr} \Rightarrow & 4 I=\int_0^{\pi / 4} \log \cos 2 x d x-\frac{\pi}{4} \log 2 & {\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]} \\ \Rightarrow & 4 I=2 I-\frac{\pi}{4} \log 2 & \quad \text { [from Eq. (iii)] } \\ \therefore & I=-\frac{\pi}{8} \log 2=\frac{\pi}{8} \log \left(\frac{1}{2}\right) \end{array}$$