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

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Aicharya · Accepted Answer

$	ext { Let } \begin{aligned} \mathrm{I} & =\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \mathrm{~d} x \ & =\int \frac{\sin x\left(1+\sin ^2 xight)}{\cos 2 x} \mathrm{~d} x \ & =\int \frac{\sin x\left(1+1-\cos ^2 xight)}{2 \cos ^2 x-1} \mathrm{~d} x \ & =\int \frac{\sin x\left(2-\cos ^2 xight)}{2 \cos ^2 x-1} \mathrm{~d} x \end{aligned}$ Put $\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}$ $	herefore \quad \mathrm{I}=-\int \frac{2-\mathrm{t}^2}{2 \mathrm{t}^2-1} \mathrm{dt}$ $\begin{aligned} & =\int \frac{t^2-2}{2 t^2-1} d t \ & =\frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} d t \ & =\frac{1}{2} \int\left(1-\frac{3}{2 t^2-1}ight) d t \ & =\frac{1}{2} \int d t-\frac{3}{2} \int \frac{d t}{(\sqrt{2} t)^2-1^2} \ & =\frac{1}{2} t-\frac{3}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \log \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}ight|+c \ & =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}ight|+\mathrm{c} \ A \quad A & =\frac{1}{2}, \mathrm{~B}=\frac{-3}{4 \sqrt{2}}, \mathrm{f}(x)=\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\end{aligned}$

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \mathrm{~d} x=\mathrm{A} \cos x+\mathrm{B} \log…

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