Explanation:
(B) : Let,
\(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x=k \log 3 \tag{i}\)
Put, \(\sin x-\cos x=t\)
\((\cos x+\sin x) d x=d t\)
Now squaring equation (i) -
\(1-\sin 2 x =t^{2}\)
\(\sin 2 x =1-t^{2}\)
When, \(\mathrm{x}=0, \mathrm{t}=-1\) and when, \(\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=0\)
\(\therefore \quad \mathrm{I} =\int_{-1}^{0} \frac{\mathrm{dt}}{9+16\left(1-\mathrm{t}^{2}\right)}\)
\(=\int_{-1}^{0} \frac{\mathrm{dt}}{25-16 \mathrm{t}^{2}}=\frac{1}{16} \int_{-1}^{0} \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^{2}-\mathrm{t}^{2}}\)
\(=\frac{1}{16} \times \frac{1}{2\left(\frac{5}{4}\right)} \log \left[\frac{\frac{5}{4}+\mathrm{t}}{\frac{5}{4}-\mathrm{t}}\right]_{-1}^{0}\)
\(=\frac{1}{40}\left[\log \left(\frac{5+4 \mathrm{t}}{5-4 \mathrm{t}}\right)\right]_{-1}^{0}=\frac{1}{40}\left[\log (1)-\log \left(\frac{1}{9}\right)\right]\)
\(=\frac{1}{40}(\log 9)=\frac{1}{40} \log 3^{2}\)
\(=\frac{2}{40} \log 3=\frac{1}{20} \log 3\)
\(\therefore\) On comparing with right hand side,
\(\mathrm{k}=\frac{1}{20}\)
