If I n -tan n x dx then I 0 - I 1 -2 I 2 - I 3 - I 8 - I 9 - I 10 equals n- N

The correct option is B ∑ _ n=1 ^ 9 tan ^ n x n ...

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Integration of Trigonometric Functions

If I n =∫ta...

Question

If $I_{n} = \int {tan}^{n} x d x$ then $I_{0} + I_{1} + 2 (I_{2} + I_{3} + . . . . + I_{8}) + I_{9} + I_{10}$ equals $(n \in N)$

1 + 8 \sum n = 1 \frac{{tan}^{n} x}{n}

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9 \sum n = 1 \frac{{tan}^{n} x}{n}

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8 \sum n = 2 \frac{{tan}^{n} x}{n + 1}

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9 \sum n = 1 \frac{{tan}^{n} x}{n + 1}

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Similar questions

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