Question

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) $\dfrac{u}{2} + \dfrac{{2{u^2}}}{3} + .... + \dfrac{{9{u^9}}}{{10}}$

Answer 1

We can take the integral ${I_n} = \int {{{\cot }^n}xdx}$ and expand the power as a product. Then we can use trigonometric identities to simplify the equation. Then we can integrate and simplify the equation. Then we can take the given expression and rearrange it by expanding the brackets. Then we can give necessary substitutions. Then by further simplification, we get the required solution.

Complete step by step solution:
We are given that ${I_n} = \int {{{\cot }^n}xdx}$
We can expand the power as follows,
$\Rightarrow {I_n} = \int {{{\cot }^{n - 2}}x\left( {{{\cot }^2}x} \right)dx}$
We know that ${\cot ^2}x = \cos e{c^2}x - 1$ . On substituting this on the integral, we get
$\Rightarrow {I_n} = \int {{{\cot }^{n - 2}}x\left( {\cos e{c^2}x - 1} \right)dx}$
On expanding the bracket, we get
$\Rightarrow {I_n} = \int {\left( {{{\cot }^{n - 2}}x.\cos e{c^2}x - {{\cot }^{n - 2}}} \right)dx}$
On expanding the integral, we get
$\Rightarrow {I_n} = \int {{{\cot }^{n - 2}}x.\cos e{c^2}xdx} - \int {{{\cot }^{n - 2}}dx}$
From the definition, we can say that ${I_{n - 2}} = \int {{{\cot }^{n - 2}}dx}$ . So, we can write the above equation as
$\Rightarrow {I_n} = \int {{{\cot }^{n - 2}}x.\cos e{c^2}xdx} - {I_{n - 2}}$
On rearranging, we get
$\Rightarrow {I_n} + {I_{n - 2}} = \int {{{\cot }^{n - 2}}x.\cos e{c^2}xdx}$
Now give substitution $u = \cot x$ ,
Then its derivative is $\dfrac{{du}}{{dx}} = \cos e{c^2}x$
$\Rightarrow du = \cos e{c^2}xdx$
Therefore, by substituting u and du in the equation, the integral will become
$\Rightarrow {I_n} + {I_{n - 2}} = \int {{u^{n - 2}}.du}$
We know that $\int {{x^n}.dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ . So, the integral will become
$\Rightarrow {I_n} + {I_{n - 2}} = \dfrac{{{u^{n - 1}}}}{{n - 1}}$…. (1)
Now we need to find the value of ${I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ..... + {I_8}} \right) + {I_9} + {I_{10}}$ .
Let $K = {I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ..... + {I_8}} \right) + {I_9} + {I_{10}}$
On grouping the terms as the sum of alternate terms, we get
$\Rightarrow K = \left( {{I_0} + {I_2}} \right) + \left( {{I_1} + {I_3}} \right) + \left( {{I_2} + {I_4}} \right) + .... + \left( {{I_7} + {I_9}} \right) + \left( {{I_8} + {I_{10}}} \right)$
On substituting equation (1), we get
$\Rightarrow K = \left( {\dfrac{{{u^1}}}{1}} \right) + \left( {\dfrac{{{u^2}}}{2}} \right) + \left( {\dfrac{{{u^3}}}{3}} \right) + .... + \left( {\dfrac{{{u^8}}}{8}} \right) + \left( {\dfrac{{{u^9}}}{9}} \right)$
On simplification and removing the brackets, we get
$\Rightarrow K = u + \dfrac{{{u^2}}}{2} + \dfrac{{{u^3}}}{3} + .... + \dfrac{{{u^8}}}{8} + \dfrac{{{u^9}}}{9}$
Therefore, the required solution is $u + \dfrac{{{u^2}}}{2} + \dfrac{{{u^3}}}{3} + .... + \dfrac{{{u^8}}}{8} + \dfrac{{{u^9}}}{9}$

So, the correct answer is option A.

Note:
We cannot find the integral of each term. We can only find a relation between 2 terms in the given series. While taking ${\cot ^2}x$ in the ${1^{st}}$ step, we must note that the remaining power is reduced and not increased. We must give the substitutions which are given in the question. While integrating, we must find the derivative of the new variable and then give the substitution to change the variable of integration. We need not to substitute the variable after integration as the answer is also in terms of the same variable.