If In=∫cotn⁡xdx then I0+I1+2I2+I3+I4+I5=

If In=∫cot2⁡xdx then In+In−2=−cotn-1xn−1n=2⇒I2+I0=-cotxn=3⇒I3+I1==−cot2⁡x2n=4=I4+I2=−cot3⁡x3n=5⇒I5+I3=−cos4⁡x4add I0+I1+2I2+I3+I4+I5 =−cot⁡x+cot2⁡x2+cot3⁡x3+cot4⁡x4+c

If In=∫cotn⁡xdx then I0+I1+2I2+I3+I4+I5=

If In=∫cot2⁡xdx then In+In−2=−cotn-1xn−1n=2⇒I2+I0=-cotxn=3⇒I3+I1==−cot2⁡x2n=4=I4+I2=−cot3⁡x3n=5⇒I5+I3=−cos4⁡x4add I0+I1+2I2+I3+I4+I5 =−cot⁡x+cot2⁡x2+cot3⁡x3+cot4⁡x4+c

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$If I_{n} = \int \cot^{n} x d x then I_{0} + I_{1} + 2 (I_{2} + I_{3}) + I_{4} + I_{5} =$

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$- [Cot x + \frac{{Cot}^{3} x}{3} + \frac{{Cot}^{5} x}{5}] + c$

$- [Cot x + \frac{{Cot}^{2} x}{2} + \frac{{Cot}^{3} x}{3} + \frac{{Cot}^{4} x}{4}] + c$

$- [Cot x + \frac{{Cot}^{3} x}{3}] + c$

$- [\frac{{Cot}^{2} x}{2} + \frac{{Cot}^{4} x}{4}] + c$

answer is A.

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Detailed Solution

If $I_{n} = \int \cot^{2} x d x$ then $I_{n} + I_{n - 2} = \frac{- \cot^{n - 1} x}{n - 1}$
$\begin{array}{l} n = 2 \Rightarrow I_{2} + I_{0} = - c o t x \\ n = 3 \Rightarrow I_{3} + I_{1} = = - \frac{\cot^{2} x}{2} \\ n = 4 = I_{4} + I_{2} = - \frac{\cot^{3} x}{3} \end{array}$
$n = 5 \Rightarrow I_{5} + I_{3}$ $= - \frac{\cos^{4} x}{4}$
add $I_{0} + I_{1} + 2 [I_{2} + I_{3}) + I_{4} + I_{5}$ $= - [\cot x + \frac{\cot^{2} x}{2} + \frac{\cot^{3} x}{3} + \frac{\cot^{4} x}{4}] + c$