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Answer:  $y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C$

Hint: To solve this we convert $cotx$  to $\frac{\cos x}{\sin x}$  formula.

Give:  $\frac{d y}{d x}+y \tan x=x^{2} \cos ^{2} x \\$

Solution:  $\frac{d y}{d x}+P y=Q \\$

\begin{aligned} &P=\tan x, Q=x^{2} \cos ^{2} x \\ \end{aligned}

$I f=e^{\int P d x} \\$

$=e^{\int \tan x d x} \\$

$=e^{\log \sec x} \\$

$=\sec x$

\begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(\sec x) d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(1 / \cos x) d x+C \\ &=y \sec x=\int x^{2} \cos x d x+C \\ &=y \sec x=x^{2} \int \cos x d x-\int(2 x \cos x d x) d x+C \end{aligned}

Using integration by parts

\begin{aligned} &=y \sec x=x^{2} \sin x-2 \int x^{2} \sin x d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C \end{aligned}

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