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If $\int \frac{1+\cos (4 x)}{\cot (x)-\tan (x)} d x=A \cos (4 x)+B$, then $A$ is equal to
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The correct answer is:
$\frac{-1}{8}$
$\int \frac{1+\cos (4 x)}{\cos x-\tan x} d x=A \cos (4 x)+B$, then $A=$ ?
Let $\begin{aligned} I & =\int \frac{1+\cos (2 \times 2 x)}{\cot x-\tan x} d x=\int \frac{1+2 \cos ^2(2 x)-1}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} d x \\ & =\int \frac{2 \cos ^2(2 x)}{\cos ^2 x-\sin ^2 x} \cdot(\sin x \cdot \cos x) d x \\ I & =2 \int \frac{\cos ^2 2 x}{\cos 2 x}(\sin x \cdot \cos x) d x\end{aligned}$
$\begin{aligned} I & =\int \cos (2 x)(\sin 2 x) d x \\ & =\frac{1}{2} \int 2 \sin (2 x) \cos (2 x) d x=\frac{1}{2} \int \sin 4 x d x \\ I & =\frac{-\cos 4 x}{8}+C=k \cos (4 x)+C \\ \therefore \quad k & =-\frac{1}{8}\end{aligned}$
Let $\begin{aligned} I & =\int \frac{1+\cos (2 \times 2 x)}{\cot x-\tan x} d x=\int \frac{1+2 \cos ^2(2 x)-1}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} d x \\ & =\int \frac{2 \cos ^2(2 x)}{\cos ^2 x-\sin ^2 x} \cdot(\sin x \cdot \cos x) d x \\ I & =2 \int \frac{\cos ^2 2 x}{\cos 2 x}(\sin x \cdot \cos x) d x\end{aligned}$
$\begin{aligned} I & =\int \cos (2 x)(\sin 2 x) d x \\ & =\frac{1}{2} \int 2 \sin (2 x) \cos (2 x) d x=\frac{1}{2} \int \sin 4 x d x \\ I & =\frac{-\cos 4 x}{8}+C=k \cos (4 x)+C \\ \therefore \quad k & =-\frac{1}{8}\end{aligned}$
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