Question
Question: If\[{I_n} = \int {{{\cot }^n}xdx} \], then \[{I_0} + {I_1} + 2\left( {{I_2} + {I_3} + ...... + {I_8}...
IfIn=∫cotnxdx, then I0+I1+2(I2+I3+......+I8)+I9+I10equals to: (whereu=cotx).
A.u+2u2+.......+9u9
B.−(u+2u2+.......+9u9)
C.−(u+2!u2+.......+9!u9)
D.2u+32u2+.......+109u2
Solution
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 In=∫cotnxdxin the reduction form and then use the chain rule to simplify and find the value ofI0+I1+2(I2+I3+......+I8)+I9+I10.
Given In=∫cotnxdx the use the reduction method, we can write
In=∫cotnxdx=∫cotn−2cot2xdx
We knowcosec2x=1+cot2x, hence we can write
SinceIn=∫cotnxdxhence we can write ∫cotn−2xdx=In−2
So we can write:
Now let us assume cotx=t
By differentiating cotx=t with respect to t we get,
Hence we can write:
In+In−2=∫cotn−2xcosec2xdx =−∫tn−2dtNow integrate:
In+In−2=−∫tn−2dt =−n−2+1tn−2+1=−n−1tn−1=−n−1cotn−1xHence,
In+In−2=−n−1cotn−1x where, n⩾2
Now replace n with (n+2), we can write
In+2+In=−n+1cotn+1x
Now we have to find the value ofI0+I1+2(I2+I3+......+I8)+I9+I10, write the given function in the pair of the common difference of 2 as In+2,In:
Since u=cotxis given hence, we can write
I0+I1+2(I2+I3+......+I8)+I9+I10=−(u+2u2+3u3+..........+9u9)
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.