First slide

Methods of integration

Question

$I_{n} = \int \cot^{n} x d x then I_{10} is equal to$

Moderate

A

$\frac{\cot^{9} x}{9} - I_{8} + C$
B

$\frac{- \cot^{9} x}{9} - I_{8} + C$
C

$\frac{- \cot^{11} x}{11} - I_{9} + C$
D

$\frac{\cot^{11} x}{11} - I_{9} + C$

Solution

$I_{10} + I_{8} = \int \cot^{10} x d x + \int \cot^{8} x d x$
$\begin{aligned} = \int \cot^{8} x (\cot^{2} x + 1) d x \\ = \int \cot^{8} x {cosec}^{2} x d x \end{aligned}$
$put \cot x = t$
$\begin{array}{l} = \int t^{8} (- d t) \\ = \frac{- t^{9}}{9} \\ = - \frac{\cot^{9} x}{9} \\ I_{10} = - \frac{\cot^{9} x}{9} - I_{8} + C \end{array}$

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