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$k \in N, \int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x=$
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The correct answer is:
$\frac{\tan x}{\sin ^k x}+C$
$I=\int \frac{1-k \cos ^2 x}{\sin ^k x \cdot \cos ^2 x} d x$
$\begin{aligned} & =\int \frac{\left(\sec ^2 x-k\right)}{\sin ^k x} d x \\ & =\int(\sin x)^{-k} \sec ^2 x d x-k \int \frac{d x}{\sin ^k x} \\ & =(\sin x)^{-k} \int \sec ^2 x d x-\end{aligned}$
$\int\left[(-k)(\sin x)^{-k-1} \cdot \cos x \int \sec ^2 x d x\right] d x-k \int \frac{d x}{\sin ^k x}$
$\begin{aligned}=(\sin x)^{-k} \tan x+ & k \int(\sin x)^{-k} \\ & \cot x \cdot \tan x d x-k \int \frac{d x}{\sin ^k x}\end{aligned}$
$=\frac{\tan x}{\sin ^k x}+k \int \frac{d x}{\sin ^k x}-k \int \frac{d x}{\sin ^k x}+C=\frac{\tan x}{\sin ^k x}+C$
$\begin{aligned} & =\int \frac{\left(\sec ^2 x-k\right)}{\sin ^k x} d x \\ & =\int(\sin x)^{-k} \sec ^2 x d x-k \int \frac{d x}{\sin ^k x} \\ & =(\sin x)^{-k} \int \sec ^2 x d x-\end{aligned}$
$\int\left[(-k)(\sin x)^{-k-1} \cdot \cos x \int \sec ^2 x d x\right] d x-k \int \frac{d x}{\sin ^k x}$
$\begin{aligned}=(\sin x)^{-k} \tan x+ & k \int(\sin x)^{-k} \\ & \cot x \cdot \tan x d x-k \int \frac{d x}{\sin ^k x}\end{aligned}$
$=\frac{\tan x}{\sin ^k x}+k \int \frac{d x}{\sin ^k x}-k \int \frac{d x}{\sin ^k x}+C=\frac{\tan x}{\sin ^k x}+C$
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