Mathematics - Indefinite Integration Question with Solution | TestHub

Let $\mathrm{I}=\int \frac{\cos 8x+1}{\cot 2x-\tan 2x}dx$ Now, ${\mathrm{D}}^r=\cot 2x-\tan 2x=\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}$ $$=\frac{{\cos }^{2}2x-{\sin }^{2}2x}{\sin 2x\cos 2x}=\frac{2\cos 4x}{\sin 4x}$$ $\begin{array}{rl}\therefore \mathrm{I}& =\int \frac{2{\cos }^{2}4x}{\frac{2\cos 4x}{\sin 4x}}dx=\int \frac{2{\cos }^{2}4x\cdot \sin 4x}{2\cos 4x}dx\\ & =\frac{1}{2}\int \sin 8xdx=-\frac{1}{2}\frac{\cos 8x}{8}+k\\ & =-\frac{1}{16}\cdot \cos 8x+k\\ & \text{ Now },-\frac{1}{16}\cdot \cos 8x+k=\mathrm{A}\cos 8x+k\\ & \Rightarrow \mathrm{A}=-\frac{1}{16}\end{array}$