Q. If $I_{n} = 0 \int \frac{π}{4} tan^{n} x d x$ where $n$ is positive integer then $I_{10} + I_{8}$ is equal to

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A

9

16%

B

$\frac{1}{7}$

26%

C

$\frac{1}{8}$

32%

D

$\frac{1}{9}$

24%

Solution:

$I_{n} + I_{n - 2} = \frac{1}{n - 1}$