If In=∫tannx dx and I0+I1+2I2+2I3+2I4+I5+I6=∑k=1ntankxk then n =

In=∫tannx dxw.k.t. ⇒ In+In−2=tann−1xn−1 →I0+I1+2I2+2I3+2I4+I5+I6 =(I2+I0)+(I3+I1)+(I4+I2)+(I5+I3)+(I6+I4) =tan⁡x+tan2⁡x2+tan3⁡x3+tan4⁡x4+tan5⁡x5=∑k=15tankxkComparing , n = 5

If In=∫tannx dx and I0+I1+2I2+2I3+2I4+I5+I6=∑k=1ntankxk then n =

In=∫tannx dxw.k.t. ⇒ In+In−2=tann−1xn−1 →I0+I1+2I2+2I3+2I4+I5+I6 =(I2+I0)+(I3+I1)+(I4+I2)+(I5+I3)+(I6+I4) =tan⁡x+tan2⁡x2+tan3⁡x3+tan4⁡x4+tan5⁡x5=∑k=15tankxkComparing , n = 5

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If $I_{n} = \int \tan^{n} x d x$ and $I_{0} + I_{1} + 2 I_{2} + 2 I_{3} + 2 I_{4} + I_{5} + I_{6} = \sum_{k = 1}^{n} \frac{\tan^{k} x}{k}$ then n =

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Detailed Solution

$I_{n} = \int \tan^{n} x d x$
w.k.t. $\Rightarrow I_{n} + I_{n - 2} = \frac{\tan^{n - 1} x}{n - 1}$
$\begin{aligned} \to I_{0} + I_{1} + 2 I_{2} + 2 I_{3} + 2 I_{4} + I_{5} + I_{6} \\ = (I_{2} + I_{0}) + (I_{3} + I_{1}) + (I_{4} + I_{2}) + (I_{5} + I_{3}) + (I_{6} + I_{4}) \\ = \tan x + \frac{\tan^{2} x}{2} + \frac{\tan^{3} x}{3} + \frac{\tan^{4} x}{4} + \frac{\tan^{5} x}{5} \end{aligned}$
$= \sum_{k = 1}^{5} \frac{\tan^{k} x}{k}$
Comparing , n = 5