\(\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \mathrm{~d} x=\mathrm{A} \cos x+\mathrm{B} \log \mathrm{f}(x)+\mathrm{c}\)
(where \(\mathrm{c}\) is a constant of integration). Then values of A, B and \(\mathrm{f}(x)\) are

\(\text { Let }\mathrm{I} =\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \mathrm{~d} x \)
\( =\int \frac{\sin x\left(1+\sin ^2 x\right)}{\cos 2 x} \mathrm{~d} x \)
\( =\int \frac{\sin x\left(1+1-\cos ^2 x\right)}{2 \cos ^2 x-1} \mathrm{~d} x \)
\( =\int \frac{\sin x\left(2-\cos ^2 x\right)}{2 \cos ^2 x-1} \mathrm{~d} x\)
Put \(\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}\)
\(
\therefore \mathrm{I}=-\int \frac{2-\mathrm{t}^2}{2 \mathrm{t}^2-1} \mathrm{dt}
\)
\(=\int \frac{t^2-2}{2 t^2-1} d t \)
\( =\frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} d t \)
\( =\frac{1}{2} \int\left(1-\frac{3}{2 t^2-1}\right) d t \)
\( =\frac{1}{2} \int d t-\frac{3}{2} \int \frac{d t}{(\sqrt{2} t)^2-1^2} \)
\( =\frac{1}{2} t-\frac{3}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \log \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+c \)
\( =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+\mathrm{c} \)
\( A A =\frac{1}{2}, \mathrm{~B}=\frac{-3}{4 \sqrt{2}}, \mathrm{f}(x)=\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\)