The correct option is B ^n+1xn+1+c I = ∫(^n+2x+^nx)dx = ∫^nx.cosec ^2xdx Put #160; z = x ⇒ #160;dz = cosec ^2 x dx ∴ I = ∫ z^ndz = ...

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Substitution Method to Remove Indeterminate Form

∫n+2x+nxdx=

Question

$\int ({cot}^{n + 2} x + {cot}^{n} x) d x =$

\frac{- {cot}^{n + 1} x}{n + 1} + c

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\frac{- {cot}^{n - 1} x}{n - 1} + C

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\frac{- {cot}^{n + 3} x}{n + 3} + c

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\frac{- {cot}^{2 n} x}{2 n} + C

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Solution

The correct option is B $\frac{- {cot}^{n + 1} x}{n + 1} + c$
$I = \int ({cot}^{n + 2} x + {cot}^{n} x) d x = \int {cot}^{n} x . {cosec}^{2} x d x$
Put $z = cot x \Rightarrow d z = - {cosec}^{2} x d x$
$∴ I = \int - z^{n} d z = - \frac{z^{n + 1}}{n + 1} + c = - \frac{{cot}^{n + 1} x}{n + 1} + c$

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{^{n} C}_{r - 1} + {^{n + 1} C}_{r - 1} + {^{n + 2} C}_{r - 1} + . . . . + {^{2 n} C}_{r - 1} = {^{2 n + 1} C}_{r^{2} - 182} - {^{n} C}_{r}

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