Question

If $\int \frac{\cos 8x+1}{\tan 2x-\cot 2x}dx$ = a cos 8x + C, then a =

(a) $-\frac{1}{16}$

(b) $\frac{1}{8}$

(c) $\frac{1}{16}$

(d) $-\frac{1}{8}$

Answer 1

(c) $\frac{1}{16}$

$$\begin{gathered} \mathrm{If}\int \left(\frac{\cos 8x+1}{\tan 2x-\cot 2x}\right)dx=a\cos 8x+C...\left(1\right) \\ \mathrm{Considering}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{eq}\left(1\right) \\ \int \left(\frac{\cos 8x+1}{\tan 2x-\cot 2x}\right)dx \\ \Rightarrow \int \left(\frac{2{\cos }^{2}4x}{{\displaystyle \frac{\sin 2x}{\cos 2x}}-{\displaystyle \frac{\cos 2x}{\sin 2x}}}\right)dx \\ \Rightarrow \int \frac{2{\cos }^{2}4x}{\left({\sin }^{2}2x-{\cos }^{2}2x\right)}\times \sin 2x\cos 2x \\ \Rightarrow \int \left[\frac{-{\cos }^{2}4x\times 2\sin 2x·\cos 2x}{{\cos }^{2}2x-{\sin }^{2}2x}\right]dx \\ \Rightarrow \int \frac{-{\cos }^{2}4x\times \sin 4x}{\cos 4x}dx\left(\because \cos 2x={\cos }^{2}x-{\sin }^{2}x\right) \\ \Rightarrow \frac{1}{2}\int -2\sin 4x\cos 4xdx \\ \Rightarrow \frac{-1}{2}\int \sin 8xdx \\ \Rightarrow -\frac{1}{2}\left[\frac{-\cos 8x}{8}\right]+C \\ =\frac{1}{16}\left[\cos 8x\right]+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}....\left(2\right) \\ \mathrm{Comparing}\mathrm{RHS}\mathrm{of}\mathrm{eq}\left(1\right)\mathrm{with}\mathrm{the}\mathrm{eq}\left(2\right) \\ \therefore a=\frac{1}{16} \end{gathered}$$