If $\int \frac{\cos 8 x+1}{\cot 2 x- an 2 x} \mathrm{~d} x=\mathrm{A} \cos 8 x+\mathrm{c}$, where $\mathrm{c}$ is an arbitrary constant, then the value of $\mathrm{A}$ is

Question

If $\int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} \mathrm{~d} x=\mathrm{A} \cos 8 x+\mathrm{c}$, where $\mathrm{c}$ is an arbitrary constant, then the value of $\mathrm{A}$ is

Aicharya · Accepted Answer

$\begin{aligned} I & =\int \frac{\cos 8 x+1}{\cot 2 x-	an 2 x} d x \ & =\int \frac{2 \cos ^2\left(\frac{8 x}{2}ight)}{\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}} d x\end{aligned}$ $\ldots\left[\because 1+\cos 	heta=2 \cos ^2 \frac{	heta}{2}ight]$ $\begin{aligned} & =\int \frac{2 \cos ^2(4 x) 	imes \sin 2 x 	imes \cos 2 x}{\cos ^2 2 x-\sin ^2 2 x} \mathrm{~d} x \ & =\int \frac{\cos ^2(4 x) \sin (4 x)}{\cos (4 x)} \mathrm{d} x \ & \quad \cdots\left[\begin{array}{r} \because \sin 2 	heta=2 \sin 	heta \cos 	heta 	ext { and } \ \cos 2 	heta=\cos ^2 	heta-\sin ^2 	heta \end{array}ight] \end{aligned}$ $\begin{aligned} & =\frac{1}{2} \int 2 \sin (4 x) \cos (4 x) \mathrm{d} x \ & =\frac{1}{2} \int \sin 8 x \mathrm{~d} x \ & =\frac{-\cos 8 x}{2 	imes 8}+c \ & =\frac{-\cos 8 x}{16}+c \end{aligned}$ Comparing with ' $\mathrm{A} \cos 8 x+\mathrm{c}$ ', we get $\mathrm{A}=\frac{-1}{16}$

If $\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \mathrm{~d} x=\mathrm{A} \cos 8 x+\mathrm{c}$, where…

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