Question

If \[{{I_n }} = \int {{{\cot }^n}{\text{x}}} {\text{ dx}}\]and \[I_{\text{0}} + I_{\text{1}}{\text{ + 2(}}I_{\text{2}} + I_{\text{3}}.....I_{\text{8}}{\text{)}} + I_{\text{9}} + I_{\text{10}} = A({\text{u }} + \dfrac{{{{\text{u}}^2}}}{2} + ... + \dfrac{{{{\text{u}}^9}}}{9}) + {\text{C}}\] where \[{\text{u}} = \cot x\] and C is an arbitrary constant, then
A) A=2
B) A=-1
C) A=1
D) A is dependent on x

Answer 1

Hint: We will apply the rules of integration on \[{{I_n }}\] and find the value of \[I_{\text{0}} + I_{\text{1}}{\text{ + 2(}}I_{\text{2}} + I_{\text{3}}.....I_{\text{8}}{\text{)}} + I_{\text{9}} + I_{\text{10}}\] and then equate it with the given value to get the value of A.

Formula Used:
The identity used is:
\[{\text{co}}{{\text{t}}^2}{\text{x}} = {\csc ^2}x - 1\]

Complete step by step solution:
We are given that:
\[{\text{I_n }} = \int {{{\cot }^{n - 2}}{\text{x}}} {\text{.co}}{{\text{t}}^2}{\text{x dx}}\]
Now we know that,
\[{\text{co}}{{\text{t}}^2}{\text{x}} = {\csc ^2}x - 1\]
Therefore, applying this formula in the above equation we get:
\[
 {\text{I_n }} = \int {{{\cot }^{n - 2}}{\text{x}}} \left( {{\text{cs}}{{\text{c}}^2}{\text{x}} - 1} \right){\text{ dx}} \\
 {\text{I_n }} = \int {{{\cot }^{n - 2}}{\text{x}}} .{\csc ^2}x{\text{ dx}} - \int {{{\cot }^{n - 2}}{\text{x}}} {\text{ dx}} \\
 \]
Now since we know that,
\[{\text{I_n - 2 }} = \int {{{\cot }^{n - 2}}{\text{x}}} {\text{ dx}}\]
Therefore replacing it in above equation we get:
\[
 {\text{I_n }} = \int {{{\cot }^{n - 2}}{\text{x}}} .{\csc ^2}x{\text{ dx}} - {\text{I_n - 2 }} \\
 {\text{I_n + I_n - 2 }} = \int {{{\cot }^{n - 2}}{\text{x}}} .{\csc ^2}x{\text{ dx}} \\
 \]
Now Let \[\cot x = {\text{t}}\]
Then, \[{\text{ - cs}}{{\text{c}}^2}x{\text{ dx}} = {\text{dt}}\]
Substituting these values in the above equation we get:
\[{\text{I_n + I_n - 2 }} = - \int {{t^{n - 2}}} {\text{dt}}\]
Now applying the following formula:
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \]
We get:-
\[{\text{I_n + I_n - 2 }} = - \dfrac{{{t^{n - 1}}}}{{n - 1}} + C\]
Now substituting back the value of t we get:-
\[{\text{I_n + I_n - 2 }} = - \dfrac{{{{\cot }^{n - 1}}x}}{{n - 1}} + C\]
Since,
\[I_{\text{0}} + I_{\text{1}}{\text{ + 2(}}I_{\text{2}} + I_{\text{3}}.....I_{\text{8}}{\text{)}} + I_{\text{9}} + I_{\text{10}} = \left( {I_{\text{2}} + I_{\text{0}}} \right) + \left( {I_{\text{3}} + I{\text{1}}} \right) + (I_4 + I_2) + (I_5 + I_3) + (I_6 + I_4) + (I_7 + I_5) + \left( {I_8 + I_6} \right) + \left( {I_9 + I_7} \right) + \left( {I_10 + I_8} \right)\]
Hence applying the above derived formula we get:
\[
 I_{\text{0}} + I_{\text{1}}{\text{ + 2(}}I_{\text{2}} + I_{\text{3}}.....I_{\text{8}}{\text{)}} + I_{\text{9}} + I_{\text{10}} = \left( {I_{\text{2}} + I_{\text{0}}} \right) + \left( {I_{\text{3}} + I_{\text{1}}} \right) + (I_4 + I_2) + (I_5 + I_3) + (I_6 + I_4) + (I_7 + I_5) + \left( {I_8 + I_6} \right) + \left( {I_9 + I_7} \right) + \left( {I_10 + I_8} \right) \\
 {\text{ = }} - \left( {\dfrac{{{{\cot }^{2 - 1}}x}}{{2 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{3 - 1}}x}}{{3 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{4 - 1}}x}}{{4 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{5 - 1}}x}}{{5 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{6 - 1}}x}}{{6 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{8 - 1}}x}}{{8 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{9 - 1}}x}}{{9 - 1}}} \right) - \left( {\dfrac{{{{\cot }^{10 - 1}}x}}{{10 - 1}}} \right) + C \\
 \]
 \[ = - \left( {\cot x + \dfrac{{{{\cot }^2}x}}{2} + \dfrac{{{{\cot }^3}x}}{3} + \dfrac{{{{\cot }^4}x}}{4}......\dfrac{{{{\cot }^9}x}}{9}} \right) + C\]
Let \[\cot x = u\]
Then putting this value in the above equation we get:-
\[I_{\text{0}} + I_{\text{1}}{\text{ + 2(}}I_{\text{2}} + I_{\text{3}}.....I_{\text{8}}{\text{)}} + I_{\text{9}} + I_{\text{10}} = - \left( {u + \dfrac{{{u^2}}}{2} + \dfrac{{{u^3}}}{3} + \dfrac{{{u^4}}}{4}......\dfrac{{{u^9}}}{9}} \right) + C\]
Comparing this with the given value we get:
$A=-1$

Hence, option B is correct.

Note:
The result of the given series can be directly applied to get the value of A. The result is:
\[{\text{I_n + I_n - 2 }} = - \dfrac{{{{\cot }^{n - 1}}x}}{{n - 1}} + C\]