¿Qué es [matemáticas] \ int _ {\ frac {\ pi} {4}} ^ {\ frac {3 \ pi} {4}} \ frac {\ sin x} {\ sin x + 1} dx [/ matemáticas] ?

Answer 1

Primero evaluamos la integral indefinida, luego conectamos los límites después.

[matemáticas] \ begin {align} \ int \ frac {\ sin x} {1+ \ sin x} \, \ mathrm dx & = \ int \ frac {\ sin x} {1+ \ sin x} \ cdot \ frac {1- \ sin x} {1- \ sin x} \, \ mathrm dx \\ & = \ int \ frac {\ sin x} {1- \ sin ^ 2 x} \, \ mathrm dx- \ int \ frac {\ sin ^ 2 x} {1- \ sin ^ 2 x} \, \ mathrm dx \\ & = – \ int \ frac {1} {\ cos ^ 2 x} \, \ mathrm d (\ cos x ) – \ int \ frac {\ sin ^ 2 x} {\ cos ^ 2 x} \, \ mathrm dx \\ & = \ frac {1} {\ cos x} + C_1- \ int \ frac {1- \ cos ^ 2 x} {\ cos ^ 2 x} \, \ mathrm dx \\ & = \ sec x + C_1- \ int \ frac {1} {\ cos ^ 2 x} \, \ mathrm dx + x + C_2 \\ & = x + \ sec x- \ int \, \ mathrm d (\ tan x) + C_1 + C_2 \\ & = x + \ sec x- \ tan x + C \ end {align} [/ math]

El siguiente paso es simplemente conectar los límites que creo que puedes tomar desde aquí.

Answer 2

[matemáticas] \ begin {align} I & = \ int \ dfrac {\ sin x} {\ sin x + 1} \ mathrm dx \\ & = \ int \ dfrac {\ sin x + 1-1} {\ sin x +1} \ mathrm dx \\ & = \ int \ left (1- \ dfrac1 {\ sin x + 1} \ right) \ mathrm dx \\ & = x- \ int \ dfrac {1- \ sin x} { \ cos ^ 2 x} \ mathrm dx \\ & = x- \ int \ sec ^ 2x \ space \ mathrm dx + \ int \ sec x \ tan x \ space \ mathrm dx \\ & = x- \ tan x + \ sec x + C \ end {align} \ tag * {} [/ math]

Conectar límites nos dará …

[matemáticas] \ begin {align} I & = x- \ tan x + \ sec x \ bigg | _ {\ frac \ pi4} ^ {\ frac {3 \ pi} 4} \\ & = \ dfrac \ pi2- \ tan \ left (\ pi- \ dfrac \ pi4 \ right) – \ sec \ dfrac \ pi4 + \ tan \ dfrac \ pi4- \ sec \ dfrac \ pi4 \\ & = \ dfrac \ pi2 + 1- \ sqrt2 + 1- \ sqrt2 \\ & = \ dfrac \ pi2 + 2-2 \ sqrt2 \ end {align} \ tag * {} [/ math]

Answer 3

Aquí vamos:

[matemáticas] \ displaystyle \ int _ {\ frac {\ pi} {4}} ^ {\ frac {3 \ pi} {4}} \ frac {\ sin \ left (x \ right)} {\ sin \ left (x \ right) +1} \ mathrm {d} x [/ math]

Primero calculamos la integral indefinida.

[matemáticas] {\ displaystyle \ int} \ dfrac {\ sin \ left (x \ right)} {\ sin \ left (x \ right) +1} \, \ mathrm {d} x [/ math]

Escriba [math] \ sin \ left (x \ right) [/ math] como [math] \ sin \ left (x \ right) +1+ \ left (1 \ right) [/ math] y divida: [math] = {\ displaystyle \ int} \ left (1- \ dfrac {1} {\ sin \ left (x \ right) +1} \ right) \ mathrm {d} x \ Rightarrow {\ displaystyle \ int} 1 \, \ mathrm {d} x – {\ displaystyle \ int} \ dfrac {1} {\ sin \ left (x \ right) +1} \, \ mathrm {d} x [/ math]

Ahora resolviendo: [matemáticas] {\ displaystyle \ int} \ dfrac {1} {\ sin \ left (x \ right) +1} \, \ mathrm {d} x [/ math]

Reescribe usando identidades trigonométricas / hiperbólicas: = [matemáticas] {\ displaystyle \ int} \ dfrac {\ sec ^ 2 \ left (\ frac {x} {2} \ right)} {\ left (\ tan \ left (\ frac {x} {2} \ right) +1 \ right) ^ 2} \, \ mathrm {d} x [/ math]

Sustituya [matemática] u = \ tan \ left (\ dfrac {x} {2} \ right) +1 \ longrightarrow \ mathrm {d} x = \ dfrac {2} {\ sec ^ 2 \ left (\ frac {x } {2} \ right)} \, \ mathrm {d} u: = \ class {steps-node} {\ cssId {steps-node-3} {2}} {\ displaystyle \ int} \ dfrac {1} {u ^ 2} \, \ mathrm {d} u [/ math]

Resolvamos ahora para: [math] {\ displaystyle \ int} \ dfrac {1} {u ^ 2} \, \ mathrm {d} u [/ math]

Aplique la regla de poder: [math] {\ displaystyle \ int} u ^ {\ mathtt {n}} \, \ mathrm {d} u = \ dfrac {u ^ {\ mathtt {n} +1}} {\ mathtt {n} +1} con \ mathtt {n} = – 2: = – \ dfrac {1} {u} [/ math]

Entonces [math] \ class {steps-node} {\ cssId {steps-node-4} {2}} {\ displaystyle \ int} \ dfrac {1} {u ^ 2} \, \ mathrm {d} u = – \ dfrac {2} {u} [/ matemáticas]

Deshacer la sustitución [matemáticas] u = \ tan \ left (\ dfrac {x} {2} \ right) +1: = – \ dfrac {2} {\ tan \ left (\ frac {x} {2} \ right) +1} [/ matemáticas]

Ahora desde [math] {\ displaystyle \ int} 1 \, \ mathrm {d} x = x [/ math] aplicando la regla constante,

[matemáticas] {\ displaystyle \ int} 1 \, \ mathrm {d} x – {\ displaystyle \ int} \ dfrac {1} {\ sin \ left (x \ right) +1} \, \ mathrm {d} x = x + \ dfrac {2} {\ tan \ left (\ frac {x} {2} \ right) +1} [/ math]

Agregar la constante: [matemáticas] {\ displaystyle \ int} \ dfrac {\ sin \ left (x \ right)} {\ sin \ left (x \ right) +1} \, \ mathrm {d} x = \ dfrac {2} {\ tan \ left (\ frac {x} {2} \ right) +1} + x + C [/ math]

Ahora, calculemos los límites.

[matemáticas] \ displaystyle \ int _ {\ frac {\ pi} {4}} ^ {\ frac {3 \ pi} {4}} \ frac {\ sin \ left (x \ right)} {\ sin \ left (x \ right) +1} \ mathrm {d} x [/ math]

Dado que [matemáticas] \ displaystyle \ int _a ^ bf \ left (x \ right) dx = F \ left (b \ right) -F \ left (a \ right) = \ lim _ {x \ to \: b-} \ left (F \ left (x \ right) \ right) – \ lim _ {x \ to \: a +} \ left (F \ left (x \ right) \ right) [/ math],

[matemáticas] \ displaystyle \ lim _ {x \ to \ frac {\ pi} {4} +} \ left (\ frac {2} {\ tan \ left (\ frac {x} {2} \ right) +1 } + x \ derecha) [/ matemáticas]

Ingrese el valor de [math] x = \ dfrac {\ pi} {4}: \ displaystyle = \ frac {2} {\ tan \ left (\ frac {\ frac {\ pi} {4}} {2} \ right) +1} + \ frac {\ pi} {4} [/ math]

Simplificando, obtenemos [math] = \ sqrt {2} + \ frac {\ pi} {4} [/ math]

[matemáticas] \ displaystyle \ lim _ {x \ to \ frac {3 \ pi} {4} -} \ left (\ frac {2} {\ tan \ left (\ frac {x} {2} \ right) + 1} + x \ derecha) [/ matemáticas]

Ingrese el valor de [math] x = \ dfrac {3 \ pi} {4}: \ displaystyle = \ frac {2} {\ tan \ left (\ frac {\ frac {3 \ pi} {4}} { 2} \ right) +1} + \ frac {3 \ pi} {4} [/ math]

Simplificando, obtenemos [matemáticas] = 2- \ sqrt {2} + \ frac {3 \ pi} {4} [/ matemáticas]

Ahora tenemos [math] = 2- \ sqrt {2} + \ frac {3 \ pi} {4} – \ left (\ sqrt {2} + \ frac {\ pi} {4} \ right) [/ math ]

Simplificando aún más, la respuesta final es [matemáticas] \ boxed {\ frac {1} {2} \ left (4-4 \ sqrt {2} + \ pi \ right)} [/ math]