If the integral $$ \int \frac{\cos 8 x+1}{\cot 2 x- an 2 x} d x=A \cos 8 x+k $$ where $k$ is an arbitrary constant, then $\mathrm{A}$ is equal to:

Question

If the integral $$ \int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} d x=A \cos 8 x+k $$ where $k$ is an arbitrary constant, then $\mathrm{A}$ is equal to:

Aicharya · Accepted Answer

Let $\mathrm{I}=\int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} d x$ Now, $\mathrm{D}^r=\cot 2 x-	an 2 x=\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}$ $ =\frac{\cos ^2 2 x-\sin ^2 2 x}{\sin 2 x \cos 2 x}=\frac{2 \cos 4 x}{\sin 4 x} $ $\begin{aligned} 	herefore \mathrm{I} &=\int \frac{2 \cos ^2 4 x}{\frac{2 \cos 4 x}{\sin 4 x}} d x=\int \frac{2 \cos ^2 4 x \cdot \sin 4 x}{2 \cos 4 x} d x \ &=\frac{1}{2} \int \sin 8 x d x=-\frac{1}{2} \frac{\cos 8 x}{8}+k \ &=-\frac{1}{16} \cdot \cos 8 x+k \ & 	ext { Now },-\frac{1}{16} \cdot \cos 8 x+k=\mathrm{A} \cos 8 x+k \ & \Rightarrow \mathrm{A}=-\frac{1}{16} \end{aligned}$

If the integral $$ \int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x=A \cos 8 x+k $$ where $k$ is an arbitrary…

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