Let $$\begin{aligned} I & =\int_0^1 x \log (1+2 x) d x \\ & =\left[\log (1+2 x) \frac{x^2}{2}\right]_0^1-\int \frac{1}{1+2 x} \cdot 2 \cdot \frac{x^2}{2} d x \\ & =\frac{1}{2}\left[x^2 \log (1+2 x)\right]_0^1-\int \frac{x^2}{1+2 x} d x \\ & =\frac{1}{2}[1 \log 3-0]-\left[\int_0^1\left(\frac{x}{2}-\frac{\frac{x}{2}}{1+2 x}\right) d x\right] \\ & =\frac{1}{2} \log 3-\frac{1}{2} \int_0^1 x d x+\frac{1}{2} \int_0^1 \frac{x}{1+2 x} d x \\ & =\frac{1}{2} \log 3-\frac{1}{2}\left[\frac{x^2}{2}\right]_0^1+\frac{1}{2} \int_0^1 \frac{1}{2}(2 x+1-1) \\ & =\frac{1}{2} \log 3-\frac{1}{2}\left[\frac{1}{2}-0\right]+\frac{1}{4} \int_0^1 d x-\frac{1}{4} \int_0^1 \frac{1}{1+2 x} d x \\ & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}[x]_0^1-\frac{1}{8}[\log |(1+2 x)|]_0^1 \\ & =\frac{1}{2} \log 3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}[\log 3-\log 1] \\ & =\frac{1}{2} \log 3-\frac{1}{8} \log 3 \\ & =\frac{3}{8} \log 3 \end{aligned}$$

Let $$I=\int_0^\pi x \log \sin x d x\quad\text{.... (i)}$$

$$\begin{aligned} I & =\int_0^\pi(\pi-x) \log \sin (\pi-x) d x \\ & =\int_0^\pi(\pi-x) \log \sin x d x \quad\text{.... (ii)}\\ 2 I & =\pi \int_0^\pi \log \sin x d x \quad\text{.... (iii)}\\ 2 I & =2 \pi \int_0^{\pi / 2} \log \sin x d x \quad \left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]\\ I & =\pi \int_0^{\pi / 2} \log \sin x d x \quad\text{.... (iv)}\\ \text{Now,}\quad I & =\pi \int_0^{\pi / 2} \log \sin (\pi / 2-x) d x\quad\text{.... (v)} \end{aligned}$$

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

$$\begin{aligned} 2 I & =\pi \int_0^{\pi / 2}(\log \sin x+\log \cos x) d x \\ 2 I & =\pi \int_0^{\pi / 2} \log \sin x \cos x d x \\ & =\pi \int_0^{\pi / 22} \log \frac{2 \sin x \cos x}{2} d x \\ 2 I & =\pi \int_0^{\pi / 2}(\log \sin 2 x-\log 2) d x \\ 2 I & =\pi \int_0^{\pi / 2} \log \sin 2 x d x-\pi \int_0^{\pi / 2} \log 2 d x \end{aligned}$$

Put $2 x=t \Rightarrow d x=\frac{1}{2} d t$

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

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

$$\begin{array}{ll} \therefore & 2 I=\frac{\pi}{2} \int_0^\pi \log \sin t d t-\frac{\pi^2}{2} \log 2 \\ \Rightarrow & 2 I=\frac{\pi}{2} \int_0^\pi \log \sin x d x-\frac{\pi^2}{2} \log 2 \\ \Rightarrow & 2 I=I-\frac{\pi^2}{2} \log 2 \quad\text{[from Eq. (iii)]}\\ \therefore & I=-\frac{\pi^2}{2} \log 2=\frac{\pi^2}{2} \log \left(\frac{1}{2}\right) \end{array}$$

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

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

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

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