First slide

Evaluation of definite integrals

Question

$If I_{n} = \int_{0}^{π / 4} \tan^{n} θ d θ for n = 1, 2, 3, \dots \dots . Then I_{n - 1} + I_{n + 1} =$

Moderate

A

0
B

1
C

$\frac{1}{n}$
D

$\frac{1}{n + 1}$

Solution

$\begin{array}{l} I_{n} + I_{n - 2} = \frac{1}{n - 1} - - - - - (1) \\ Put n = n + 1 \\ I_{n + 1} + I_{n - 1} = \frac{1}{n} \end{array}$

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