# Explain how evaluate limit of integral definite on [0;pi/6] (sin x)^2n/cos x?what rule is applied here?

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### 1 Answer

Using the Mean Value Theorem `int_0^(pi/6) ((sin x)^(2n)dx)/cos x` = `(pi/6 - 0)*f(c_n)`

`` `c_n` `in (0 ; pi/6)` => `0 lt c_n lt pi/6`

The sine function is increasing on `(0;pi/6)` =>`sin 0 lt sin c_n lt sin pi/6 =gt 0 lt sin c_n lt 1/2` (a)

Raising to 2n power yields `0 lt (sin c_n)^(2n) lt(1/2)^(2n)`

The cosine function is decreasing on `(0;pi/6) ` => `cos 0gt cosc_ngtcos pi/6 =gt1gtcos c_n gt sqrt 3/2 =gt 1 lt 1/cos c_n lt 2/sqrt 3` (b)

Multiply the inequalities (a),(b):

`0 lt ((sin c_n)^(2n))/(cos c_n) lt(2/sqrt 3)*((1/2)^(2n))`

Multiply by `pi/6` (`pi/6gt0` => the direction of the inequality is not reversed):

`0 lt (pi/6)*((sin c_n)^(2n))/(cos c_n) lt (pi/6)*(2/sqrt 3)*((1/2)^(2n))`

Evaluate the limit:

`lim_(n-gtoo) 0lt lim_(n-gtoo) (pi/6)*((sin c_n)^(2n))/(cos c_n) lt lim_(n-gtoo) (pi/6)*(2/sqrt 3)*((1/2)^(2n))`

**Using squeeze limit:`lim_(n-gtoo) (pi/6)*(2/sqrt 3)*((1/2)^(2n)) = 0` =>`lim_(n-gtoo) ` `int_0^(pi/6)f(c_n)dx` = 0**