I. $\int_{-1}^1 x|x| d x=0 \quad[\because x|x|$ is an odd function $]$
II. Let $I=\int_0^{\pi / 2}\left[1+\left(\log \frac{4+3 \sin x}{4+3 \cos x}\right)\right] d x$
$$
\begin{aligned}
& I=\int_0^{\pi / 2}\left(1+\left[\log \frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right]\right) d x \\
& I=\int_0^{\pi / 2}\left(1+\left(\log \frac{4+3 \cos x}{4+3 \sin x}\right)\right) d x \\
& \Rightarrow \quad 2 I=\int_0^{\pi / 2}\left(2+\log \frac{4+3 \sin x}{4+3 \cos x}\right. \\
& \left.+\log \frac{4+3 \cos x}{4+3 \sin x}\right) d x \\
& \Rightarrow \quad 2 I=\int_0^{\pi / 2}\left(2+\log \frac{(4+3 \sin x)(4+3 \cos x)}{(4+3 \cos x)(4+3 \sin x)}\right) d x \\
& \Rightarrow \quad Z I=\int_0^{\pi / 2}(2+\log 1) d x=\int_0^{\pi / 2} 2 d x=\pi \\
& \Rightarrow \quad I=\frac{\pi}{2} \\
&
\end{aligned}
$$
III. $\int_0^a f(x) d x=\int_0^a f(a-x) d x$



$$
\begin{aligned}
\text { IV. } \int_{-a}^a f(x) d x & =\int_0^a f(x) d x+\int_0^a f(-x) d x \\
& =\int_0^a[f(x)+f(-x)] d x
\end{aligned}
$$