# If${I_n} = \int {{{\cot }^n}xdx}$, then ${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}} \right) + {I_9} + {I_{10}}$equals to: (where$u = \cot x$).A.$u + \dfrac{{{u^2}}}{2} + ....... + \dfrac{{{u^9}}}{9}$B.$- \left( {u + \dfrac{{{u^2}}}{2} + ....... + \dfrac{{{u^9}}}{9}} \right)$C.$- \left( {u + \dfrac{{{u^2}}}{{2!}} + ....... + \dfrac{{{u^9}}}{{9!}}} \right)$D.$\dfrac{u}{2} + \dfrac{{2{u^2}}}{3} + ....... + \dfrac{{9{u^2}}}{{10}}$

Verified
146.7k+ views
Hint: Use integration reduction method which relies on recurrence relations. This method is used when the expression contains integer parameters in the form of power of elementary functions. This method can be derived from any common method of integration, like partial integration or integration by substitution.

Complete step by step solution:
In this question write ${I_n} = \int {{{\cot }^n}xdx}$in the reduction form and then use the chain rule to simplify and find the value of${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}} \right) + {I_9} + {I_{10}}$.
Given ${I_n} = \int {{{\cot }^n}xdx}$ the use the reduction method, we can write
${I_n} = \int {{{\cot }^n}xdx} = \int {{{\cot }^{n - 2}}{{\cot }^2}xdx}$
We know$\cos e{c^2}x = 1 + {\cot ^2}x$, hence we can write
${I_n} = \int {{{\cot }^{n - 2}}x\left( {\cos e{c^2}x - 1} \right)dx} \\ {I_n} = \int {{{\cot }^{n - 2}}x\cos e{c^2}xdx - \int {{{\cot }^{n - 2}}x} dx} \\$
Since${I_n} = \int {{{\cot }^n}xdx}$hence we can write $\int {{{\cot }^{n - 2}}xdx = {I_{n - 2}}}$
So we can write:
${I_n} = \int {{{\cot }^{n - 2}}x\cos e{c^2}xdx - {I_{n - 2}}} \\ {I_n} + {I_{n - 2}} = \int {{{\cot }^{n - 2}}x\cos e{c^2}xdx} \\$
Now let us assume $\cot x = t$
By differentiating $\cot x = t$ with respect to t we get,
$\dfrac{d}{{dx}}(\cot x) = \dfrac{{dt}}{{dx}} \\ - \cos e{c^2}xdx = dt \\$
Hence we can write:
${I_n} + {I_{n - 2}} = \int {{{\cot }^{n - 2}}x\cos e{c^2}xdx} \\ = - \int {{t^{n - 2}}dt} \\$
Now integrate:
${I_n} + {I_{n - 2}} = - \int {{t^{n - 2}}dt} \\ = - \dfrac{{{t^{n - 2 + 1}}}}{{n - 2 + 1}} = - \dfrac{{{t^{n - 1}}}}{{n - 1}} = - \dfrac{{{{\cot }^{n - 1}}x}}{{n - 1}} \\$
Hence,
${I_n} + {I_{n - 2}} = - \dfrac{{{{\cot }^{n - 1}}x}}{{n - 1}}$ where, $n \geqslant 2$
Now replace n with (n+2), we can write
${I_{n + 2}} + {I_n} = - \dfrac{{{{\cot }^{n + 1}}x}}{{n + 1}}$
Now we have to find the value of${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}} \right) + {I_9} + {I_{10}}$, write the given function in the pair of the common difference of 2 as ${I_{n + 2}},{I_n}$:
${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}} \right) + {I_9} + {I_{10}} = \left( {{I_0} + {I_2}} \right) + \left( {{I_1} + {I_3}} \right) + \left( {{I_2} + {I_4}} \right) + \left( {{I_3} + {I_5}} \right) + \left( {{I_4} + {I_6}} \right) + \left( {{I_5} + {I_7}} \right) + \left( {{I_8} + {I_{10}}} \right) + \left( {{I_7} + {I_9}} \right) \\ = - \cot x - \dfrac{{{{\cot }^2}x}}{2} - \dfrac{{{{\cot }^3}x}}{3} - .......... - \dfrac{{{{\cot }^9}x}}{9} \\ = - \left( {\cot x + \dfrac{{{{\cot }^2}x}}{2} + \dfrac{{{{\cot }^3}x}}{3} + .......... + \dfrac{{{{\cot }^9}x}}{9}} \right) \\$
Since $u = \cot x$is given hence, we can write
${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}} \right) + {I_9} + {I_{10}} = - \left( {u + \dfrac{{{u^2}}}{2} + \dfrac{{{u^3}}}{3} + .......... + \dfrac{{{u^9}}}{9}} \right)$
Hence option B is correct.

Note: While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.