Let $I=\int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$
We know that,
$$
\begin{array}{rlrl}
(\sin x-\cos x)^2= & \sin ^2 x+\cos ^2 x-2 \sin x \cos x \\
\Rightarrow & (\sin x-\cos x)^2 & =1-2 \sin 2 x \\
\Rightarrow & & \sin 2 x & =1-(\sin x-\cos x)^2 \\
\text { Put } & & \sin x-\cos x & =t
\end{array}
$$
Put $\quad \sin x-\cos x=t$
On differentiating Eq. (i) w.r.t $x$,
$$
\begin{aligned}
& (\cos x+\sin x) d x=d t \\
& \Rightarrow \quad(\cos x+\sin x)=\frac{d t}{d x} \\
& \Rightarrow \quad d x=\frac{d t}{(\cos x+\sin x)} \\
& \text { When } x=0, t=0-\cos 0=-1 \\
& \text { and } \quad x=\frac{\pi}{4}, t=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0 \\
& \therefore \quad I=\int_1^0 \frac{\sin x+\cos x}{7+9\left(1-t^2\right)} \frac{d t}{(\sin x+\cos x)} \\
& =\int_1^0 \frac{1}{7+9\left(1-t^2\right)} d t \\
&
\end{aligned}
$$
$=\int_1^0 \frac{d t}{16-9 t^2}=\frac{1}{9} \int_1^0 \frac{d t}{\left(\frac{4}{3}\right)^2-t^2}$

$$
\begin{aligned}
& =\frac{1}{9} \cdot \frac{1}{2 \cdot \frac{4}{3}}\left[\log \left|\frac{\frac{4}{3}+t}{\frac{4}{3}-t}\right|\right]_1^0 \\
& =\frac{1}{24}\left[\log \left|\frac{4+3 t}{4-3 t}\right|\right]_1^0=\frac{1}{24}\left[\log 1-\log \frac{1}{7}\right] \\
& =\frac{1}{24}[\log 1-(\log 1-\log 7)]=\frac{1}{24} \log 7
\end{aligned}
$$