find-0-pi-2-sinx-ln-1-x-dx-

Question Number 49939 by turbo msup by abdo last updated on 12/Dec/18

f i n d \int_{0}^{\frac{π}{2}} s i n x l n (1 + x) d x

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

\int_{0}^{\frac{π}{2}} s i n x l n (1 + x) d x \approx (0.2 \times 0.2) \times n o s s m a l l e s t s q u a r e

Commented by mathmax by abdo last updated on 03/Nov/19

l e t d e t e r m i n e a a p p r o x i m a t e v a l u e l e t A = \int_{0}^{\frac{π}{2}} s i n x l n (1 + x) d x

b y p a r t s A = {[- c o s x l n (1 + x)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (- c o s x) \frac{d x}{1 + x}

= \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + x} d x w e h a v e c o s x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{(2 n)!}

= 1 - \frac{x^{2}}{2} + \frac{x^{2}}{24} - \Rightarrow 1 - \frac{x^{2}}{2} ⩽ c o s x ⩽ 1 \Rightarrow \frac{1 - \frac{x^{2}}{2}}{x + 1} ⩽ \frac{c o s x}{x + 1} ⩽ \frac{1}{x + 1} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{2 - x^{2}}{2 (x + 1)} d x ⩽ \int_{0}^{\frac{π}{2}} \frac{c o s x}{x + 1} d x ⩽ \int_{0}^{1} \frac{d x}{x + 1}

\int_{0}^{1} \frac{d x}{x + 1} = {[l n (x + 1)]}_{0}^{1} = l n (2)

\int_{0}^{\frac{π}{2}} \frac{2 - x^{2}}{2 (x + 1)} d x =_{x + 1 = t} \int_{1}^{1 + \frac{π}{2}} \frac{2 - {(t - 1)}^{2}}{2 t} d t

= \int_{1}^{1 + \frac{π}{2}} \frac{2 - t^{2} + 2 t - 1}{t} d t = \int_{1}^{1 + \frac{π}{2}} \frac{1 - t^{2} + 2 t}{t} d t

= \int_{1}^{1 + \frac{π}{2}} \frac{d t}{t} - \int_{1}^{1 + \frac{π}{2}} t d t + π = l n (1 + \frac{π}{2}) - {[\frac{t^{2}}{2}]}_{1}^{1 + \frac{π}{2}} + π

= l n (1 + \frac{π}{2}) - \frac{1}{2} ({(1 + \frac{π}{2})}^{2} - 1) + π \Rightarrow

l n (1 + \frac{π}{2}) - \frac{1}{2} (π + \frac{π^{2}}{4}) + π ⩽ A ⩽ l n (2) \Rightarrow

l n (1 + \frac{π}{2}) + \frac{π}{2} - \frac{π^{2}}{8} ⩽ A ⩽ l n (2) a n d w e c a n t a k e

v_{0} = \frac{1}{2} {l n (1 + \frac{π}{2}) + \frac{π}{2} - \frac{π^{2}}{8} + l n (2)} a s a p o r o x i m a t e v a l u e f o r A

Commented by mathmax by abdo last updated on 03/Nov/19

e r r o r o f t y p o a t l i n e 8

\dots = \int_{1}^{1 + \frac{π}{2}} \frac{2 - t^{2} + 2 t - 1}{2 t} d t = \int_{1}^{1 + \frac{π}{2}} \frac{1 - t^{2} + 2 t}{2 t} d t

= \int_{1}^{1 + \frac{π}{2}} \frac{d t}{2 t} - \int_{1}^{1 + \frac{π}{2}} \frac{t}{2} d t + \int_{1}^{1 + \frac{π}{2}} d t

= \frac{1}{2} l n (1 + \frac{π}{2}) - \frac{1}{4} {{(1 + \frac{π}{2})}^{2} - 1} + \frac{π}{2} = \dots . .