Question

If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]

Options

• \[- \frac{1}{16}\]

• \[\frac{1}{8}\]

• \[\frac{1}{16}\]

• \[- \frac{1}{8}\]

Answer 1

\[\frac{1}{16}\]

 

\[\text{If }\int\left( \frac{\cos 8x + 1}{\tan 2x - \cot 2x} \right)dx = a \cos 8x + C ............(1) \]
\[\text{Considering the LHS of eq. } (1)\]
\[\int\left( \frac{\cos 8x + 1}{\tan 2x - \cot 2x} \right)dx\]
\[ \Rightarrow \int\left( \frac{2 \cos^2 4x}{\frac{\sin 2x}{\cos 2x} - \frac{\cos 2x}{\sin 2x}} \right)dx\]
\[ \Rightarrow \int\frac{2 \cos^2 4x}{\left( \sin^2 2x - \cos^2 2x \right)} \times \sin 2x \cos 2x\]
\[ \Rightarrow \int\left[ \frac{- \cos^2 4x \times 2 \sin 2x \cdot \cos 2x}{\cos^2 2x - \sin^2 2x} \right]dx\]
\[ \Rightarrow \int\frac{- \cos^2 4x \times \sin 4x}{\cos 4x}dx ................\left( \because \cos 2x = \cos^2 x - \sin^2 x \right)\]
\[ \Rightarrow \frac{1}{2}\int - 2 \sin 4x \cos 4x dx \]
\[ \Rightarrow \frac{- 1}{2}\int\sin 8x dx\]
\[ \Rightarrow - \frac{1}{2}\left[ \frac{- \cos 8x}{8} \right] + C\]
\[ = \frac{1}{16}\left[ \cos 8x \right] + C ...............(2) \]
\[\text{Comparing RHS of eq. (1) with the eq. } (2)\]
\[ \therefore a = \frac{1}{16}\]