For any integer n geq 2,if I_n = int cot^n x | vedclass

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(A) We are given $I_n = \int \cot^n x \, dx$. For $n=5$,$I_5 = \int \cot^5 x \, dx$. We can write $\cot^5 x = \cot^3 x \cdot \cot^2 x = \cot^3 x (\csc

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$\frac{-\cot^4 x}{4} + \frac{\cot^2 x}{2} + \log |\sin x| + c$

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(A) We are given $I_n = \int \cot^n x \, dx$. For $n=5$,$I_5 = \int \cot^5 x \, dx$. We can write $\cot^5 x = \cot^3 x \cdot \cot^2 x = \cot^3 x (\csc^2 x - 1) = \cot^3 x \csc^2 x - \cot^3 x$. So,$I_5 = \int \cot^3 x \csc^2 x \, dx - \int \cot^3 x \, dx$. Let $u = \cot x$,then $du = -\csc^2 x \, dx$,so $\int \cot^3 x \csc^2 x \, dx = -\int u^3 \, du = -\frac{u^4}{4} = -\frac{\cot^4 x}{4}$. Now,for $\int \cot^3 x \, dx = \int \cot x (\csc^2 x - 1) \, dx = \int \cot x \csc^2 x \, dx - \int \cot x \, dx$. $\int \cot x \csc^2 x \, dx = -\frac{\cot^2 x}{2}$ and $\int \cot x \, dx = \log |\sin x|$. Thus,$\int \cot^3 x \, dx = -\frac{\cot^2 x}{2} - \log |\sin x|$. Substituting these back,$I_5 = -\frac{\cot^4 x}{4} - (-\frac{\cot^2 x}{2} - \log |\sin x|) + c = -\frac{\cot^4 x}{4} + \frac{\cot^2 x}{2} + \log |\sin x| + c$.

For any integer $n \geq 2$,if $I_n = \int \cot^n x \, dx$,then $I_5 =$

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