Mejor respuesta
Usamos la fórmula cos (A + B) = Cos A cos B – sin A sin B, para obtener la fórmula de cos 2x.
Entonces, Cos 2x = Cos (x + x) = Cos x cos x – sin x sin x = cos ^ 2 x – sin ^ 2 x
Entonces, Cos 2x = cos ^ 2 x – sin ^ 2 x … ……. (1)
Nuevamente, cos ^ 2 x – sin ^ 2 x = cos ^ 2 x – (1- cos ^ 2 x) = 2 cos ^ 2 x – 1
Entonces, Cos 2x = 2 cos ^ 2 x – 1 …………… .. (2)
De nuevo, 2 cos ^ 2 x – 1 = 2 (1 – sin ^ 2 x) – 1 = 1 – 2 sin ^ 2 x
Entonces , Cos 2x = 1 – 2 sin ^ 2 x
Por lo tanto, Cos 2x = cos ^ 2 x – sin ^ 2 x = 2 cos ^ 2 x – 1 = 1 – 2 sin ^ 2 x
Respuesta h 2>
Esto es ab * tch:
{\ Displaystyle \ int} \ dfrac {2x} {\ sin \ left (2x \ right)} \, \ mathrm {d} x
Reescribimos esta ecuación como = {\ displaystyle \ int} 2x \ csc \ left (2x \ right) \, \ mathrm {d} x \ Rightarrow \ class {steps-node} {\ cssId {steps- node-1} {2}} {\ displaystyle \ int} x \ csc \ left (2x \ right) \, \ mathrm {d} x
Resolvamos para {\ displaystyle \ int} x \ csc \ left (2x \ right) \, \ mathrm {d} x
Reescritura usando exponenciales: = {\ displaystyle \ int} \ dfrac {2 \ mathrm {i} x} {\ mathrm { e} ^ {2 \ mathrm {i} x} – \ mathrm {e} ^ {- 2 \ mathrm {i} x}} \, \ mathrm {d} x \ Rightarrow \ class {pasos-nodo} {\ cssId {pasos-nodo-2} {2 \ mathrm {i}}} {\ cdot} {\ Displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x} – \ mathrm {e} ^ {- 2 \ mathrm {i} x}} \, \ mathrm {d} x
Ahora resolviendo: {\ displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x} – \ mathrm {e} ^ {- 2 \ mathrm {i} x}} \, \ mathrm {d} x
Ponga términos sobre un denominador común: = {\ Displaystyle \ int} \ dfrac {x \ mathrm {e} ^ {2 \ mathrm {i} x}} {\ mathrm {e} ^ {4 \ mathrm {i} x} -1} \, \ mathrm {d} x
Desde \ dfrac {\ mathrm {d}} {\ mathrm {d} x} \ left [\ mathrm {i} x \ right] = \ mathrm {i}.mathtt {g}
\ mathtt {f} \: = \ ln \ left (v \ right) \ to \ mathtt {f} «\: = \ dfrac {1} {v}; \ mathtt {g} «\: = \ dfrac {1} {v + 1} \ to \ mathtt {g} \: = \ ln \ left (v + 1 \ right)
= \ ln \ left (v \ derecha) \ ln \ izquierda (v + 1 \ derecha) – {\ Displaystyle \ int} \ dfrac {\ ln \ left (v + 1 \ right)} {v} \, \ mathrm {d} v
{\ Displaystyle \ int} \ dfrac {\ ln \ left (v + 1 \ right)} {v} \, \ mathrm {d} v
Ahora resolviendo: {\ Displaystyle \ int} \ dfrac {\ ln \ left (v + 1 \ right)} {v} \, \ mathrm {d} v
Sustituya w = -v \ longrightarrow \ mathrm {d} v = – \ mathrm {d} w
= – {\ Displaystyle \ int} – \ dfrac {\ ln \ left (1-w \ right)} {w} \, \ mathrm {d} w
De nuevo, este es un dilogaritmo como el anterior: = \ operatorname {Li} \_2 \ left (w \ right)
Entonces – {\ displaystyle \ int} – \ dfrac {\ ln \ left (1-w \ right)} {w} \, \ mathrm {d} w = – \ operatorname {Li} \_2 \ left (w \ right)
Deshacer la sustitución w = -v: = – \ operatorname {Li} \_2 \ left (-v \ right)
Inserta integrales resueltas: \ ln \ left (v \ right) \ ln \ left (v + 1 \ right) – { \ Displaystyle \ int} \ dfrac {\ ln \ left (v + 1 \ right)} {v} \, \ mathrm {d} v = \ ln \ left (v \ right) \ ln \ left (v + 1 \ derecha) + \ operatorname {Li } \_2 \ left (-v \ right)
Ahora resolviendo para: {\ displaystyle \ int} \ dfrac {\ ln \ left (v \ right)} {v-1} \, \ mathrm { d} v
Sustituya w = v-1 \ longrightarrow \ mathrm {d} v = \ mathrm {d} w
= {\ displaystyle \ int} \ dfrac {\ ln \ left (w + 1 \ right)} {w} \, \ mathrm {d} w
De nuevo, este es un dilogaritmo: = – \ operatorname {Li} \_2 \ left (-w \ right )
Y dado que w = v-1: = – \ operatorname {Li} \_2 \ left (1-v \ right)
Inserta integrales resueltas: – \ class {pasos -nodo} {\ cssId {pasos-nodo-8} {\ dfrac {1} {2}}} {\ Displaystyle \ int} \ dfrac {v \ ln \ left (v \ right)} {v ^ 2 + 1 } \, \ mathrm {d} v + \ class {pasos-nodo} {\ cssId {pasos-nodo-9} {\ dfrac {1} {4}}} {\ Displaystyle \ int} \ dfrac {\ ln \ left (v \ right)} {v + 1} \, \ mathrm {d} v + \ class {pasos-nodo} {\ cssId {pasos-nodo-10} {\ dfrac {1} {4}}} {\ Displaystyle \ int} \ dfrac {\ ln \ left (v \ right)} {v-1} \, \ mathrm {d} v = – \ dfrac {\ ln \ left (v \ right) \ ln \ left (v ^ 2 + 1 \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ left (-v ^ 2 \ right)} {8} + \ dfrac {\ ln \ left (v \ right) \ ln \ left (v + 1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left (-v \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ left (1-v \derecho )} {4}
Deshaga la sustitución de v = \ mathrm {e} ^ u, use \ ln \ left (\ mathrm {e} ^ u \ right) = u:
= – \ dfrac {u \ ln \ left (\ mathrm {e} ^ {2u} +1 \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {2u} \ right)} {8} + \ dfrac {u \ ln \ left (\ mathrm {e} ^ u + 1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ u \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ u \ right)} {4}
Vuelve a introducir integrales resueltas: – {\ displaystyle \ int} \ dfrac {u \ mathrm {e} ^ {2u}} {\ mathrm {e} ^ {4u} -1} \, \ mathrm {d} u = \ dfrac {u \ ln \ left (\ mathrm {e} ^ {2u} +1 \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {2u} \ derecha)} {8} – \ dfrac {u \ ln \ left (\ mathrm {e} ^ u + 1 \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ left (- \ mathrm {e } ^ u \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ u \ right)} {4}
Dado que u = \ mathrm {i} x: = \ dfrac {\ mathrm {i} x \ ln \ left (\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right)} {4} + \ dfrac {\ nombre de operador {Li} \_2 \ left (- \ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {8} – \ dfrac {\ mathrm {i} x \ ln \ left (\ mathrm {e } ^ {\ mathrm {i} x} +1 \ right)} {4} – \ dfrac {\ operatorname {Li} \_2 \ l eft (- \ mathrm {e} ^ {\ mathrm {i} x} \ right)} {4} + \ dfrac {\ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ {\ mathrm {i } x} \ right)} {4}
Inserte integrales resueltas: \ class {steps-node} {\ cssId {steps-node-12} {2 \ mathrm {i}}} {\ cdot} {\ Displaystyle \ int} \ dfrac {x} {\ mathrm {e} ^ {2 \ mathrm {i} x} – \ mathrm {e} ^ {- 2 \ mathrm {i} x}} \, \ mathrm {d} x = – \ dfrac {x \ ln \ left (\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right)} {2} + \ dfrac {\ mathrm {i} \ nombre de operador {Li} \_2 \ left (- \ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {4} + \ dfrac {x \ ln \ left (\ mathrm {e} ^ {\ mathrm {i} x} +1 \ right)} {2} – \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {\ mathrm {i} x} \ right) } {2} + \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)} {2}
Inserta integrales resueltas: \ class {steps-node} {\ cssId {steps-node-13} {2}} {\ displaystyle \ int} x \ csc \ left (2x \ right) \, \ mathrm {d } x = -x \ ln \ left (\ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right) + \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {2} + x \ ln \ left (\ mathrm {e} ^ {\ mathrm {i} x} +1 \ right) – \ mathrm {i} \ nombre del operador {Li} \_2 \ left (- \ mathrm {e} ^ { \ mathrm {i} x} \ right) + \ mathrm {i} \ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ {\ mathrm {i} x} \ right)
Todo lo que tenemos que hacer ahora es aplicar la función de valor absoluto a los argumentos de las funciones logaritmos para extender el dominio de la antiderivada:
{\ displaystyle \ int} 2x \ csc \ left (2x \ right ) \, \ mathrm {d} x = -x \ ln \ left (\ left | \ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right | \ right) + x \ ln \ left ( \ left | \ mathrm {e} ^ {\ mathrm {i} x} +1 \ right | \ right) + \ dfrac {\ mathrm {i} \ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {2 \ mathrm {i} x} \ right)} {2} – \ mathrm {i} \ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {\ mathrm {i} x} \ right) + \ mathrm {i} \ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ {\ mathrm {i} x} \ right) + C \ Rightarrow \ boxed {\ dfrac {\ mathrm {i} \ left (\ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {2 \ mathrm {i} x} \ right) -2 \ operatorname {Li} \_2 \ left (- \ mathrm {e} ^ {\ mathrm {i} x} \ right) +2 \ operatorname {Li} \_2 \ left (1- \ mathrm {e} ^ {\ mathrm {i} x} \ right) \ right)} {2} -x \ left ( \ ln \ left (\ left | \ mathrm {e} ^ {2 \ mathrm {i} x} +1 \ right | \ right) – \ ln \ left (\ left | \ mathrm {e} ^ {\ mathrm { i} x} +1 \ derecha | \ derecha) \ derecha) + C}