First slide

Methods of integration

Question

If $I_{n} = \int \cot^{n} x d x$ , and $I_{0} + I_{1} + 2 (I_{2} + \dots + I_{8})$
$+ I_{9} + I_{10} = A (u + \frac{u^{2}}{2} + \dots + \frac{u^{9}}{9})$ , where $u = \cot x$ then

Moderate

A

$A = 1$
B

$A = - 1$
C

$A = 1 / 2$
D

$A = - 1 / 2$

Solution

$I_{n} + I_{n + 2} = \int \cot^{n} x {cosec}^{2} x d x = - \frac{1}{n + 1} \cot^{n + 1} x$
Now ,
$\begin{array}{l} I_{0} + I_{1} + 2 (I_{2} + I_{3} + \dots + I_{8}) + I_{9} + I_{10} \\ = (I_{0} + I_{2}) + (I_{1} + I_{3}) + (I_{2} + I_{4}) \\ + \dots + (I_{7} + I_{9}) + (I_{8} + I_{10}) \\ = - (u + \frac{1}{2} u^{2} + \frac{1}{3} u^{3} + \dots + \frac{1}{9} u^{9}) \end{array}$
where , $u = \cot x$
Thus , $A = - 1$

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