find ∫_0 ^(π/2) sinx ln(1+x) dx


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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
$let determine a approximate value let A=∫_0 ^(π/2) sinx ln(1+x)dx by parts A =[−cosx ln(1+x)]_0 ^(π/2) −∫_0 ^(π/2) (−cosx)(dx/(1+x)) =∫_0 ^(π/2) ((cosx)/(1+x))dx we have cosx =Σ_(n=0) ^∞ (((−1)^n x^(2n) )/((2n)!)) =1−(x^2 /2) +(x^2 /(24))− ⇒1−(x^2 /2)≤cosx ≤1 ⇒((1−(x^2 /2))/(x+1))≤((cosx)/(x+1))≤(1/(x+1)) ⇒ ∫_0 ^(π/2) ((2−x^2 )/(2(x+1)))dx ≤∫_0 ^(π/2) ((cosx)/(x+1))dx ≤∫_0 ^1 (dx/(x+1)) ∫_0 ^1 (dx/(x+1)) =[ln(x+1)]_0 ^1 =ln(2) ∫_0 ^(π/2) ((2−x^2 )/(2(x+1)))dx =_(x+1=t) ∫_1 ^(1+(π/2)) ((2−(t−1)^2 )/(2t))dt =∫_1 ^(1+(π/2)) ((2−t^2 +2t−1)/t)dt =∫_1 ^(1+(π/2)) ((1−t^2 +2t)/t)dt =∫_1 ^(1+(π/2)) (dt/t) −∫_1 ^(1+(π/2)) tdt +π =ln(1+(π/2))−[(t^2 /2)]_1 ^(1+(π/2)) +π =ln(1+(π/2))−(1/2)((1+(π/2))^2 −1)+π ⇒ ln(1+(π/2))−(1/2)(π+(π^2 /4))+π≤A≤ln(2) ⇒ ln(1+(π/2))+(π/2)−(π^2 /8) ≤A ≤ln(2) and we can take v_0 =(1/2){ ln(1+(π/2))+(π/2)−(π^2 /8) +ln(2)} as aporoximate value for A$
$l e t d e t e r m i n e a a p p r o x i m a t e v a l u e l e t A = \int_{0}^{\frac{π}{2}} s i n x l n (1 + x) d x$ $b y p a r t s A = {[- c o s x l n (1 + x)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} (- c o s x) \frac{d x}{1 + x}$ $= \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + x} d x w e h a v e c o s x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{(2 n)!}$ $= 1 - \frac{x^{2}}{2} + \frac{x^{2}}{24} - \Rightarrow 1 - \frac{x^{2}}{2} ⩽ c o s x ⩽ 1 \Rightarrow \frac{1 - \frac{x^{2}}{2}}{x + 1} ⩽ \frac{c o s x}{x + 1} ⩽ \frac{1}{x + 1} \Rightarrow$ $\int_{0}^{\frac{π}{2}} \frac{2 - x^{2}}{2 (x + 1)} d x ⩽ \int_{0}^{\frac{π}{2}} \frac{c o s x}{x + 1} d x ⩽ \int_{0}^{1} \frac{d x}{x + 1}$ $\int_{0}^{1} \frac{d x}{x + 1} = {[l n (x + 1)]}_{0}^{1} = l n (2)$ $\int_{0}^{\frac{π}{2}} \frac{2 - x^{2}}{2 (x + 1)} d x =_{x + 1 = t} \int_{1}^{1 + \frac{π}{2}} \frac{2 - {(t - 1)}^{2}}{2 t} d t$ $= \int_{1}^{1 + \frac{π}{2}} \frac{2 - t^{2} + 2 t - 1}{t} d t = \int_{1}^{1 + \frac{π}{2}} \frac{1 - t^{2} + 2 t}{t} d t$ $= \int_{1}^{1 + \frac{π}{2}} \frac{d t}{t} - \int_{1}^{1 + \frac{π}{2}} t d t + π = l n (1 + \frac{π}{2}) - {[\frac{t^{2}}{2}]}_{1}^{1 + \frac{π}{2}} + π$ $= l n (1 + \frac{π}{2}) - \frac{1}{2} ({(1 + \frac{π}{2})}^{2} - 1) + π \Rightarrow$ $l n (1 + \frac{π}{2}) - \frac{1}{2} (π + \frac{π^{2}}{4}) + π ⩽ A ⩽ l n (2) \Rightarrow$ $l n (1 + \frac{π}{2}) + \frac{π}{2} - \frac{π^{2}}{8} ⩽ A ⩽ l n (2) a n d w e c a n t a k e$ $v_{0} = \frac{1}{2} {l n (1 + \frac{π}{2}) + \frac{π}{2} - \frac{π^{2}}{8} + l n (2)} a s a p o r o x i m a t e v a l u e f o r A$
Commented by mathmax by abdo last updated on 03/Nov/19
$error of typo at line 8 ...=∫_1 ^(1+(π/2)) ((2−t^2 +2t−1)/(2t))dt =∫_1 ^(1+(π/2)) ((1−t^2 +2t)/(2t))dt =∫_1 ^(1+(π/2)) (dt/(2t))−∫_1 ^(1+(π/2)) (t/2)dt +∫_1 ^(1+(π/2)) dt =(1/2)ln(1+(π/2))−(1/4){ (1+(π/2))^2 −1}+(π/2) =.....$
$e r r o r o f t y p o a t l i n e 8$ $. . . = \int_{1}^{1 + \frac{π}{2}} \frac{2 - t^{2} + 2 t - 1}{2 t} d t = \int_{1}^{1 + \frac{π}{2}} \frac{1 - t^{2} + 2 t}{2 t} d t$ $= \int_{1}^{1 + \frac{π}{2}} \frac{d t}{2 t} - \int_{1}^{1 + \frac{π}{2}} \frac{t}{2} d t + \int_{1}^{1 + \frac{π}{2}} d t$ $= \frac{1}{2} l n (1 + \frac{π}{2}) - \frac{1}{4} {{(1 + \frac{π}{2})}^{2} - 1} + \frac{π}{2} = . . . . .$
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