Question-25300

Question Number 25300 by SAGARSTARK last updated on 07/Dec/17

Commented by prakash jain last updated on 08/Dec/17

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Commented by moxhix last updated on 08/Dec/17

{(t a n x)}^{2} = 1 (x = π / 4)

{(t a n x)}^{2} < 1 (0 ⩽ x < π / 4)

∴ \underset{n \to \infty}{l i m} {(t a n x)}^{2 n} = {\begin{cases} 1 (x = π / 4) \\ 0 (0 ⩽ x < π / 4) \end{cases}

Double subscripts: use braces to clarify

= \int_{0}^{π / 4} 0 d x = 0

a_{n} = \int_{0}^{π / 4} {t a n}^{2 n} x d x

= \int_{0}^{π / 4} {t a n}^{2 n - 2} x (\frac{1}{{c o s}^{2} x} - 1) d x

= \int_{0}^{π / 4} {t a n}^{2 n - 2} x (\frac{1}{{c o s}^{2} x}) d x - a_{n - 1}

t = t a n x, d t = \frac{1}{{c o s}^{2} x} d x, x : [0, π / 4] \Rightarrow t : [0, 1]

= \int_{0}^{1} t^{2 n - 2} d t - a_{n - 1}

a_{n} = \frac{1}{2 n - 1} - a_{n - 1}

∴ \frac{1}{2 n - 1} = a_{n} + a_{n - 1}

a_{1} = \int_{0}^{π / 4} {t a n}^{2} x d x = {[t a n x - x]}_{0}^{π / 4} = 1 - \frac{π}{4}

\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k - 1}}{2 k - 1} = 1 + \underset{k = 2}{\sum^{n}} {(- 1)}^{k - 1} (a_{k} + a_{k - 1})

= 1 + {- (a_{2} + a_{1}) + (a_{3} + a_{2}) - \dots + {(- 1)}^{n - 1} (a_{n} + a_{n - 1}) [

= 1 + (- a_{1} + {(- 1)}^{n - 1} a_{n})

\underset{(n \to \infty)}{\to} 1 - a_{1} + 0 = \frac{π}{4}

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