t-0-pi-2-sin-x-t-cos-x-2-dx-find-the-value-of-the-extermum-of-t-

Question Number 165581 by mnjuly1970 last updated on 04/Feb/22

φ (t) = \int_{0}^{\frac{π}{2}} {(s i n (x) + t c o s (x))}^{2} d x

f i n d t h e v a l u e o f t h e e x t e r m u m

o f φ (t) .

Answered by aleks041103 last updated on 04/Feb/22

s i n (x) + t c o s (x) = \sqrt{1 + t^{2}} s i n (x + a r c t a n (t))

\Rightarrow φ (t) = \int_{0}^{π / 2} \frac{{s i n}^{2} (x + a)}{1 + t^{2}} d x =

= (1 + t^{2}) \int_{a}^{a + π / 2} {s i n}^{2} x d x, a = {t a n}^{- 1} t

\int {s i n}^{2} x d x = \int \frac{1 - c o s (2 x)}{2} d x =

= \frac{x}{2} - \frac{1}{4} s i n (2 x)

\Rightarrow \int_{a}^{a + π / 2} {s i n}^{2} x d x = \frac{π}{4} - \frac{1}{4} (s i n (π + 2 a) - s i n (2 a)) =

= \frac{π}{4} + s i n (a) c o s (a) =

= \frac{π}{4} + t a n (a) {c o s}^{2} (a) =

= \frac{π}{4} + \frac{t}{1 + {t a n}^{2} (a)} = \frac{π}{4} + \frac{t}{1 + t^{2}}

\Rightarrow φ (t) = \frac{π}{4} (1 + t^{2}) + t

Commented by aleks041103 last updated on 04/Feb/22

φ (t) = \frac{π}{4} t^{2} + t + \frac{π}{4}

φ^{'} = \frac{π}{2} t + 1 = 0

\Rightarrow t = - \frac{2}{π}

\Rightarrow φ_{e x t r} = \frac{π}{4} \frac{4}{π^{2}} - \frac{2}{π} + \frac{π}{4} =

= - \frac{1}{π} + \frac{π}{4} = \frac{π^{2} - 4}{4 π} = φ_{e x t r .}

Answered by mahdipoor last updated on 05/Feb/22

φ^{^{'}} (t) = \lim_{k \to t} \frac{φ (k) - φ (t)}{x - t} =

\lim_{k \to t} \frac{\int_{0}^{π / 2} (k^{2} - t^{2}) {c o s}^{2} x + 2 (k - t) s i n x . c o s x}{k - t} =

\lim_{k \to t} \int_{0}^{π / 2} (k + t) {c o s}^{2} x + 2 s i n x . c o s x =

2 t \int_{0}^{π / 2} {c o s}^{2} x + \int_{0}^{π / 2} 2 s i n x . c o s x = \frac{π}{2} t + 1

\Rightarrow φ^{^{'}} (t) = 0 \Rightarrow t = - \frac{2}{π}

\Rightarrow φ (t) = \int φ^{^{'}} (t) . d t = \frac{π}{4} t^{2} + t + c

\Rightarrow φ (0) = \int_{0}^{π / 2} {(s i n x)}^{2} = \frac{π}{4} \Rightarrow c = \frac{π}{4}

m i n φ = φ (- \frac{2}{π}) = \frac{π^{2} - 4}{4 π}

Commented by aleks041103 last updated on 04/Feb/22

O n t h e s e c o n d l i n e :

\frac{\int_{0}^{π / 2} [(k^{2} - t^{2}) {c o s}^{2} x + 2 (k - t) s i n x . c o s x] d x}{k - t}

s i n c e

φ (t) = \int_{0}^{π / 2} {(s i n (x) + t c o s (x))}^{2} d x

Commented by mahdipoor last updated on 04/Feb/22

t h a n k s f o r y o u r h i n t

Commented by mahdipoor last updated on 05/Feb/22

b o z o r g v a r i a z i z

Commented by mnjuly1970 last updated on 04/Feb/22

a l i b o o d j e n a b e m a h d i p o o r .

m a m n o o n .