If the integral ∫ cos 8 x+1 cot 2 x-tan 2 x d x=A cos 8 x+k where k is an arbitrary constant, then A is equal to:

Let $\displaystyle \mathrm{I}=\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x$I=∫cot2x−tan2xcos8x+1​dx Now, $\mathrm{D}^r=\cot 2 x-\tan 2 x=\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}$Dr=cot2x−tan2x=sin2xcos2x​−cos2xsin2x​ $$=\frac{\cos ^2 2 x-\sin ^2 2 x}{\sin 2 x \cos 2 x}=\frac{2 \cos 4 x}{\sin 4 x}$$=sin2xcos2xcos22x−sin22x​=sin4x2cos4x​ $\displaystyle \begin{aligned} \therefore \mathrm{I} &=\int \frac{2 \cos ^2 4 x}{\frac{2 \cos 4 x}{\sin 4 x}} d x=\int \frac{2 \cos ^2 4 x \cdot \sin 4 x}{2 \cos 4 x} d x \\ &=\frac{1}{2} \int \sin 8 x d x=-\frac{1}{2} \frac{\cos 8 x}{8}+k \\ &=-\frac{1}{16} \cdot \cos 8 x+k \\ & \text { Now },-\frac{1}{16} \cdot \cos 8 x+k=\mathrm{A} \cos 8 x+k \\ & \Rightarrow \mathrm{A}=-\frac{1}{16} \end{aligned}$∴I​=∫sin4x2cos4x​2cos24x​dx=∫2cos4x2cos24x⋅sin4x​dx=21​∫sin8xdx=−21​8cos8x​+k=−161​⋅cos8x+k Now ,−161​⋅cos8x+k=Acos8x+k⇒A=−161​​