Answer
423.9k+ 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)