? ? + c 2 2 x z 2 2 x – a Encuentre una formula directa para la aplicación de la siguiente integral dx x 2 – a 2 x 2 – a 2 ? x = a sec z dx x 2 – a 2 ? sec z dz = = ? a sec z tg z dz (a tg z ) x = a sec z x 2 = a 2 sec2 z si x = a sec z ? dx = a sec z tg z dz Tabla de integrales ? sec z dz = ln sec z + tg z + c Reemplazando x 2 – a 2 = x 2 – a 2 = x 2 – a 2 = a 2 sec 2 z – a 2 a 2 (sec 2 z – 1) a 2 (tg 2 z ) ? sec z dz = Ln sec z + tg z + c x 2 – a 2 = a tg z ? sec z dz = Ln x a + x 2 – a 2 a + c sec z = x a cos z = a x ? sec z dz = Ln x + x 2 – a 2 a ? sec z dz = Ln x + x – a – Ln a + c Cancelando términos semejantes Pero c1 = – Ln a + c x 2 – a 2 ? sec z dz = Ln x + x – a + c1 a dx 2 2 ? 2 2 = Ln x + x – a + c 8
? x ? ? ? = ? dz ? 1 a ? ? ? ? ? ? 1 1 a 1 ? x ? x a ? ? ? x ? ? Encuentre una formula directa para la aplicación de la siguiente integral x dx x x 2 – a 2 dx x 2 – a 2 = 1 a arc sec ? ? + c donde a > 0 ? a ? x 2 – a 2 ? x = a sec z dx x x 2 – a 2 = ? a sec z tg z dz 1 (a sec z ) a tg z a x 2 = a 2 sec 2 z Si x = a sec z ? dx = a sec z tg z dz ? dz = Tabla de integrales x 2 – a 2 = x 2 – a 2 = a 2 sec 2 z – a 2 a 2 ? sec 2 z – 1? ? dz = z + c x 2 – a 2 = a 2 ? tg 2 ? z ? ? ? dz = (z ) + c a Reemplazando (z ) + c = 1 arc sec ? x ? + c a a ? a ? x 2 – a 2 = a tg z si x = a sec z ? sec z = z = arc sec ? ? ? a ? dx x x 2 – a 2 = 1 a arc sec ? ? + c ? a ? x x 2 – a 2 z a 9
2 2 2 ? ? 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? Encuentre una formula directa para la aplicación de la siguiente integral ? a + x dx ? a + x 2 dx = ? a sec z ? a sec 2 z dz ? = ? a 2 sec 3 z dz ? ? a 2 + x 2 x = a tg z ? x = a tg z a 2 ? sec3 z dz = Se resuelve la integral a 2 ? sec3 z dz por partes ? u dv = u * v – ? v du a 2 ? sec 3 z dz = a 2 ? sec 2 z * sec z dz x 2 = a 2 tg 2 z si x = a tg z ? dx = a sec 2 z dz a 2 + x 2 = a 2 + a 2 tg 2 z a 2 + x 2 = a 2 (1 + tg 2 z ) Se resuelve por partes u = sec z du = sec z tg z dz dv = sec2 z ? dv = ? sec z dz a 2 + x 2 = a 2 ? sec 2 z ? ? ? a 2 + x 2 = a sec z v = tg z ? u dv = u * v – ? v du a 2 ? sec 2 z * sec z dz = a 2 [sec z * tg z – ? (tg z )* sec z * tgz dz] a 2 ?sec z * tgz – ? tg 2 z * sec z dz? Reemplazando la Identidad trigonométrica tg2 z = sec2 z – 1 a 2 ?sec z * tg z – ? ? sec 2 – 1? * sec z dz? a 2 ?sec z * tg z – ? sec3 z dz + ? sec z dz? a 2 ? sec 3 z dz = a 2 ?sec z * tg z – ? sec 3 z dz + ? sec z dz? a 2 ? sec 3 z dz = a 2 sec z * tg z – a 2 ? sec 3 z dz + a 2 ? sec z dz Ordenando como una ecuación cualquiera y simplificando los términos semejantes a 2 ? sec 3 z dz + a 2 ? sec 3 z dz = a 2 sec z * tg z + a 2 ? sec z dz 2a 2 ? sec 3 z dz = a 2 sec z * tg z + a 2 ? sec z dz Dividiendo la ecuación por 2 10
a 2 a 2 a x a 2 2 ? ? sec z = + a x ? ? 2 2 ? a 2 ? 2 ? ? + ? ? 2 2 1 ? 2 ? a 2 2 ? + ? – a 2 ? sec3 z dz = a 2 2 sec z * tg z + a 2 2 ? sec z dz Tabla de integrales ? a ? sec z dz = ln sec z + tg z + c a 2 ? sec 3 z dz = sec z * tg z + Ln sec z + tg z + c 2 2 2 2 2 + x 2 dx = a sec z * tg z + a Ln sec z + tg z + c 2 2 si x = a tg z ? tg z = cos z = a 2 + x 2 ? a + x dx = a 2 ? a 2 + x 2 2 ? a ? ? ? x ? a 2 ? * ? a ? + 2 Ln ? a 2 + x 2 a + c a 2 + x 2 = a sec z a 2 + x 2 a ? a + x dx = a 2 ? x a 2 + x 2 2 ? ? ? a 2 ? + 2 Ln ? a 2 + x 2 + x a + c ? a 2 + x 2 dx = 1 ? x a 2 + x 2 ? ? a 2 ? 2 Ln a 2 + x 2 + x a + c Propiedad de los logaritmos C1 = – Ln a + c ? a + x dx = ? x a 2 + x 2 ? ? a 2 ? 2 Ln a 2 + x 2 + x a 2 2 Ln a + c a 2 + x 2 z x a ? a 2 2 + x 2 dx = x a 2 + x 2 + a Ln 2 2 a 2 + x 2 + x + C1 11
2 2 2 2 = ? ? ? ? ? ? ? ? ? ? ? ? 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? Encuentre una formula directa para la aplicación de la siguiente integral ? x 2 – a 2 dx ? x – a dx = ? a tg z (a sec z tg z dz) = ? a sec z tg z dz x 2 – a 2 x = a sec z ? x = a sec z a 2 ? tg 2 z sec z dz = Identidades trigonometricas x 2 = a 2 sec 2 z si x = a sec z ? dx = a sec z tg z dz tg 2 z = sec 2 z – 1 a 2 ? tg 2 z sec z dz = a 2 ? ? sec 2 z – 1? sec z dz x 2 – a 2 x 2 – a 2 = a 2 sec 2 z – a 2 a 2 (sec 2 z – 1) a 2 ? ? sec 2 z – 1? sec z dz = a 2 ? sec 2 z sec z dz – a 2 ? sec z dz a 2 ? sec3 z dz – a 2 ? sec z dz = Se resuelve la integral a 2 ? sec 3 z dz por partes ? u dv = u * v – ? v du a 2 ? sec3 z dz = a 2 ? sec 2 z * sec z dz Se resuelve por partes x 2 – a 2 = a 2 ? tg 2 z ? x 2 – a 2 = a tg z u = sec z du = sec z tg z dz dv = sec2 z ? dv = ? sec z dz v = tg z ? u dv = u * v – ? v du a 2 ? sec 2 z * sec z dz = a 2 [sec z * tg z – ? (tg z )* sec z * tgz dz] a 2 ?sec z * tgz – ? tg 2 z * sec z dz? Reemplazando la Identidad trigonométrica tg2 z = sec2 z – 1 a 2 ?sec z * tg z – ? ? sec 2 – 1? * sec z dz? a 2 ?sec z * tg z – ? sec3 z dz + ? sec z dz? a 2 ? sec3 z dz = a 2 ?sec z * tg z – ? sec3 z dz + ? sec z dz? 12
x a a 2 ? ? 2 ? 2 ? ? ? a x ? ? 2 2 2 x ? ? ? a * ? ? ? ? z a 2 2 ? a 2 ? – ? 2 2 x 2 a 2 ? sec 3 z dz = a 2 sec z * tg z – a 2 ? sec 3 z dz + a 2 ? sec z dz Ordenando como una ecuación cualquiera y simplificando los términos semejantes a 2 ? sec 3 z dz + a 2 ? sec 3 z dz = a 2 sec z * tg z + a 2 ? sec z dz 2a 2 ? sec 3 z dz = a 2 sec z * tg z + a 2 ? sec z dz Dividiendo la ecuación por 2 a 2 ? sec3 z dz = a 2 2 sec z * tg z + a 2 2 ? sec z dz Tabla de integrales ? sec z dz = ln sec z + tg z + c Regresando a la integral inicial después de resolver a 2 ? sec 3 z dz por partes. x = a sec z si x = a sec z ? sec z = a 2 ? sec3 z dz – ? a 2 ? sec z dz ? = sec z * tg z + a 2 2 ? sec z dz – ? a ? sec z dz ? cos z = a 2 2 sec z * tg z + a 2 2 ? sec z dz – a ? sec z dz x 2 – a 2 = a tg z Se reducen términos semejantes a 2 2 sec z * tg z – a 2 2 ? sec z dz tg z = x 2 – a 2 a aplicando la tabla de integrales a 2 2 sec z * tg z – a 2 2 Ln sec z + tg z + c Reemplazando ? x – a dz = a 2 2 sec z * tg z – a 2 2 Ln sec z + tg z + c x 2 – a 2 x 2 – a 2 dz = a 2 ? x ? ? x 2 – a 2 2 ? a ? ? ? a 2 x ? – 2 Ln a + ? x 2 – a 2 a + c ? x – a dz = a 2 ? x x 2 – a 2 2 ? ? ? a 2 ? 2 ? Ln x + x 2 – a 2 a + c ? x – a dz = x 2 – a 2 – a 2 2 Ln x + x 2 – a 2 a + c 13
? x 2 2 x 2 Propiedad de los logaritmos 2 2 2 – a 2 dz = x x 2 – a 2 – a Ln x + x 2 – a 2 + a Ln a + c 2 2 2 Pero: C1 = a 2 2 Ln a + c ? x – a dx = x 2 – a 2 + a 2 2 Ln x 2 – a 2 + x + C1 14
2 2 ? a 1 2 1 2 ? ? ? ? a 2 ? 1 ? ? ? ? ? ? a 2 ? a 2 ? 4 ? ? ? x a ? x ? ? a 2 ? 4 ? ? 2 ? 2 ? a ? a ?? ? ? ? a 2 x a 2 ? x * a 2 – x 2 2 ? 2 a ? ? ? ? ? 1 ? 2 ? ? ? x 2 2 x x 2 Encuentre una formula directa para la aplicación de la siguiente integral ? a – x dx 2 – x 2 dx = ? a cos z * (a cos z dz) = ? a 2 cos 2 z dz Reemplazando la Identidad trigonométrica a 2 – x 2 ? x = a sen z cos 2 z = (1 + cos2z ) x 2 = a 2 sen 2 z Si x = asen z ? dx = a cos z dz a 2 ? cos 2 z dz = a 2 ? 2 (1 + cos 2z ) dz = a ? (1 + cos 2z ) dz 2 a 2 – x 2 = a 2 – a 2 sen 2 z a 2 2 ? dz + a 2 2 ? cos 2z dz a 2 – x 2 = a 2 ?1 – sen 2 z ? a 2 2 z + ? sen 2z ? + c 2 ? 2 ? a 2 – x 2 = a 2 ? cos 2 z ? z + ? sen 2z ? + c 2 Reemplazando la Identidad trigonométrica a 2 – x 2 = a cos z si x = a sen z ? sen z = sen 2 z = 2 sen z cos z z = arc sen ? ? ? a ? cos z = a 2 – x 2 a a 2 2 z + ? ? ? sen 2z ? + c = ? a 2 2 z + a 2 4 (2 sen z cos z ) + c tg z = x a 2 – x 2 Reemplazando a 2 ? ? x ?? a 2 ? x arc sec? ?? + * arc sec + ? ? a 2 a 2 – x 2 ? a ? ? ? ? + c = ? a 2 2 arc sec x a + ? x a 2 – x 2 ? ? + c ? a z a 2 – x 2 ? a – x = a 2 2 arc sen + a a 2 – x 2 – + c 15
? ? = ? 1 ? 1 1 dz ? ? ? ? 1 1 a a a x 1 – 1 a dx 1 ? Encuentre una formula directa para la aplicación de la siguiente integral dx x x 2 + a 2 dx ? x x 2 + a 2 1 sec z dz a tg z a sec 2 z dz a tg z (a sec z ) x 2 + a 2 ? x = a tg z x = a tg z x2 = a2 tg2 z si x = a tg z ? dx = a sec2 z dz dz 1 cos z a sen z = ? a sen z x 2 + a 2 = a 2 tg 2 z + a 2 cos z 1 ? csc z dz a Tabla de integrales ? csc z dz = ln csc z – ctg z + c x 2 + a 2 = a 2 ? tg 2 z + 1? ? ? x 2 + a 2 = a 2 ? sec 2 z ? ? ? x 2 + a 2 = a sec z ? csc z dz = Ln csc z – ctg z + c Reemplazando x 2 + a 2 a Ln csc z – ctg z + c = Ln a x x + c si csc z = x = a tg z ? tg z = x 2 + a 2 x x a ctg z = x 2 + a 2 x x x = Ln 2 + a 2 a x 2 + a 2 – a x + c z a 16
? ? ? ? ? ? ?? ? ? ? ? ? x a 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? Encuentre una formula directa para la aplicación de la siguiente integral x 2 dx x 2 + a 2 x 2 + a 2 ? x = a tg z x 2 dx x 2 + a 2 = ? a 2 tg 2 z ? a sec 2 z dz ? ? ? a sec z x = a tg z x2 = a2 tg2 z a 2 ? tg 2 z sec z dz Identidad trigonometrica tg2 z = sec2 z – 1 a 2 ? tg 2 z sec z dz = a 2 ? ? sec 2 z – 1? (sec z dz ) a 2 ? sec3 z dz – a 2 ? sec z dz si x = a tg z ? dx = a sec 2 z dz x 2 + a 2 = a 2 tg 2 z + a 2 x 2 + a 2 = a 2 ? tg 2 z + 1? ? ? x 2 + a 2 = a 2 ? sec 2 z ? ? ? x 2 + a 2 = a sec z x 2 dx x 2 + a 2 = a 2 ? sec3 z dz – a 2 ? sec z dz si x = a tg z ? tg z = Esta solución se resuelve primero la integral a 2 ? sec3 z dz por partes y después la otra integral ? u dv = u * v – ? v du a 2 ? sec3 z dz = a 2 ? sec 2 z * sec z dz Se resuelve por partes u = sec z du = sec z tg z dz dv = sec2 z ? dv = ? sec z dz v = tg z ? u dv = u * v – ? v du a 2 ? sec3 z dz = a 2 ? sec z * sec z dz = a 2 [sec z * tg z – ? (tg z )* sec z * tgz dz] a 2 ? sec3 z dz = a 2 ?sec z * tgz – ? tg 2 z * sec z dz? Reemplazando la Identidad trigonométrica tg2 z = sec2 z – 1 a 2 ?sec z * tg z – ? ? sec 2 z – 1? * sec z dz? = a 2 ?sec z * tg z – ? sec3 z dz + ? sec z dz? 17
? ? ? ? x a x 2 dx a x 2 a 2 2 a 2 2 a 2 1 2 2 x a – ? ? * – + a x a 2 ? sec3 z dz = a 2 ?sec z * tg z – ? sec3 z dz + ? sec z dz? a 2 ? sec3 z dz = a 2 sec z * tg z – a 2 ? sec3 z dz + a 2 ? sec z dz ordenando como una ecuación cualquiera y reduciendo términos semejantes a 2 ? sec3 z dz + a 2 ? sec3 z dz = a 2 sec z * tg z + a 2 ? sec z dz 2a 2 ? sec3 z dz = a 2 sec z * tg z + a 2 ? sec z dz Dividiendo la ecuación por 2 se obtiene la primera parte de la solución a 2 ? sec3 z dz = a 2 2 sec z * tg z + a 2 2 ? sec z dz si x = a tg z ? tg z = La integral inicial es : ? = a 2 ? sec3 z dz – a 2 ? sec z dz x 2 + a 2 cos z = ctg z = a x 2 + a 2 sec z = x 2 + a 2 a Se reemplaza en la integral inicial y se sigue resolviendo la integral a 2 ? sec3 z dz – a 2 ? sec z dz = sec z * tg z + ? sec z dz – a ? sec z dz Reduciendo términos semejantes a 2 ? sec3 z dz – a 2 ? sec z dz = sec z * tg z – a 2 ? sec z dz + c Tabla de integrales ? sec z dz = ln sec z + tg z + c x 2 + a 2 z 2 a 2 ? sec3 z dz – a 2 ? sec z dz = a sec z * tg z 2 Reemplazando a 2 2 Ln sec z + tg z + c a 2 ? sec3 z dz – a 2 ? sec z dz = a 2 2 sec z * tg z – a 2 2 Ln sec z + tg z + c a 2 ? sec 3 z dz – a 2 ? sec z dz = a 2 ? x 2 + a 2 2 ? a ? ? x a 2 ? a 2 ? Ln x 2 + a 2 a + c 18
? a 2 ? – ? x 2 ? x 2 + 1 2 ? x 2 a 2 ? sec 3 z dz – a 2 ? sec z dz = a 2 ? x x 2 + a 2 2 ? ? ? a 2 ? 2 ? Ln x 2 + a 2 + x a + c a 2 ? sec 3 z dz – a 2 ? sec z dz = x 2 + a 2 – a 2 2 Ln x 2 + a 2 + x a + c x 2 dx x 2 + a 2 = a 2 ? sec 3 z dz – a 2 ? sec z dz = x 2 + a 2 – a 2 2 Ln x 2 + a 2 + x a 2 2 Ln a + c Pero: C1 = 2 a Ln a + c x 2 dx x 2 + a 2 = a 2 ? sec3 z dz – a 2 ? sec z dz = x 2 + a 2 – a 2 2 Ln x 2 + a 2 + x + c1 19
? ? dz 2 ?? ?? 1 2 1 2 2 1 1 x x a ? Encuentre una formula directa para la aplicación de la siguiente integral dx x 2 a 2 – x 2 a 2 – x 2 ? x = a sen z dx x 2 a 2 – x 2 1 1 ? a 2 sen 2 z = ? a cos z dz a 2 sen 2 z (a cos z) x = a sen z x 2 = a 2 sen 2 z si x = a sen z ? dx = a cos z dz 1 2 ? csc z dz a 2 Tabla de integrales ? csc z dz = – ctg z + c a 2 – x 2 = a 2 – x 2 = a 2 – x 2 = a 2 – a 2 sen 2 z a 2 ?1 – sen 2 z ? ? ? a 2 ? cos 2 z ? ? ? a 2 – x 2 = a cos z ? csc z dz = – a a Reemplazando – (ctg z) + c = – ( a 2 a 2 ctg z + c a 2 – x 2 x ) + c si x = a sen z ? sen z = tg z = ctg z = a 2 – x 2 a 2 – x 2 x dx x 2 a 2 – x 2 = – a 2 – x 2 a 2 x + c a x z a 2 – x 2 20
? x ? ? 2 z ? ? ? ?? ? 2 ? 2 ? ? ? ? 2 ? 2 ? ? ? x ? x a 2 ? x ? ? Encuentre una formula directa para la aplicación de la siguiente integral a 2 – x 2 x 2 dx a 2 – x 2 ? x = a sen z a 2 – x 2 dx 2 a 2 cos 2 z dz a 2 sen 2 z = ? (a cos z) (a cos z dz ) a 2 sen 2 z x = a sen z x 2 = a 2 sen 2 z si x = a sen z ? dx = a cos z dz a 2 – x 2 = a 2 – a 2 sen 2 z ? ctg z dz Identidad trigonometrica ctg2 z = csc2 z – 1 a 2 – x 2 = a 2 – x 2 = a 2 ?1 – sen 2 z ? ? ? a 2 ? cos 2 ? ? ctg z dz = ? ? csc z – 1? dz ? ? csc z – 1? dz = ? csc z dz – ? dz Tabla de integrales a 2 – x 2 = a cos z si x = a sen z ? sen z = z = arc sen ? ? ? a ? ? csc z dz = – ctg z + c tg z = x a 2 – x 2 ctg z = a 2 – x 2 x Reemplazando ? csc 2 z dz – ? dz = – ctg z – z + c a 2 – x 2 x 2 dx = – a 2 – x 2 x – arc sen ? ? + c ? a ? a x z a 2 – x 2 21
? ? ? ? ? ? ? x 2 dx ? ? 3 2 a 3 cos 3 z 2 3 2 3 2 = ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? 2 ? 2 ? 2 ? ? 2 Tabla de integrales ? a 2 – x 2 ? 3 2 ? ? ? ? ? ? ? ?? 3 2 x 3 2 2 x a ? ? ? x x ? ? ? ? ? Encuentre una formula directa para la aplicación de la siguiente integral x 2 dx ? a 2 – x 2 ? 3 2 ? a 2 – x 2 ? ? ? 3 2 ? x = a sen z ? a 2 sen 2 z ? (a cos z dz ) ? = ? ? a 2 – x 2 ? ? ? ? tg z dz Identidad trigonometrica sen 2 z dz cos 2 z x 2 = a 2 sen 2 z si x = a sen z ? dx = a cos z dz ? a 2 – x 2 ? = ? a 2 – a 2 sen 2 z ? ? ? ? ? tg2 z = sec2 z – 1 ? a 2 – x 2 ? ? ? 3 2 = ?a 2 ?1 – sen 2 z ?? 3 2 ? tg z dz = ? ? sec z – 1 ? dz = ? sec z dz – ? dz Tabla de integrales ? sec z dz = tg z + c ? dz = z + c ? sec z dz – ? dz = tg z – z + c ? ? = ?a 2 ? cos 2 z ?? ? a 2 – x 2 ? = a 3 cos 3 z ? ? si x = a sen z ? sen z = a z = arc sen Reemplazando tg z – z + c = a 2 – x 2 – arc sen + c a tg z = x a 2 – x 2 x 2 dx ? a 2 – x 2 ? 3 2 = x a 2 – x 2 – arc sen x a + c a x z a 2 – x 2 22
? ? 2 1 4 2 ? ? ? ? 1 2 1 x 2 1 4 – x 2 – INTEGRALES POR SUSTITUCIONES TRIGONOMETRICAS Ejercicios 7.3 Pag 571 Leythold Problema # 1 dx x 2 4 – x 2 dx x 2 4 – x 2 1 dz ? 4 sen 2 z = ? 2 cos z dz 4 sen 2 z (2 cos z ) a 2 – x 2 ? x = a sen z 4 – x 2 = 2 2 – x 2 2 2 – x 2 ? x = 2 sen z x = 2 sen z x 2 = 4 sen 2 z 1 1 ? 4 sen 2 z dz = ? csc z dz si x = 2 sen z ? dx = 2 cos z dz 4 – x 2 = 4 – 4 sen 2 z Tabla de integrales ? csc z dz = – ctg z + c 4 – x 2 = 4 – x 2 = 4 ?1 – sen 2 z ? ? ? 4 ? cos 2 z ? ? ? ? csc z dz = – ctg z + c 4 4 Reemplazando 4 – x 2 = 2 cos z si x = 2 sen z ? sen z = 1 4 (ctg z) + c = – ( 4 x ) + c tg z = x 4 – x 2 ctg z = 4 – x 2 x ? dx x 2 4 – x 2 = – 4 – x 2 4 x + c 2 x z 4 – x 2 23
? ? ? 2 ? ? ? ? 2 ? 2 ? ? ? ? 2 ? 2 ? ? z ? ? ? x 2 ? 2 ? x ? x 2 2 ? x ? ? ? ? ? ? ? 2 ? ? ? ? ? ? Ejercicios 7.3 Pag 571 Leythold Problema # 2 4 – x 2 dx x 2 a 2 – x 2 ? x = a sen z 4 – x 2 = 2 2 – x 2 4 – x 2 dx x 2 cos 2 z dz sen 2 z = ? 2 cos z (2 cos z dz ) 4 sen 2 z 2 2 – x 2 ? x = 2 sen z x = 2 sen z x 2 = 4 sen 2 z si x = 2 sen z ? dx = 2 cos z dz ? ctg z dz Identidad trigonométrica ctg2 z = csc2 z – 1 4 – x 2 = 4 – x 2 = 4 – 4 sen 2 z 4 ?1 – sen 2 z ? ? ctg z dz = ? ? csc z – 1? dz ? ? csc z – 1? dz = ? csc z dz – ? 1dz Tabla de integrales 4 – x 2 = 4 ? cos 2 ? 4 – x 2 = 2 cos z si x = 2 sen z ? sen z = ? dz = z + c ? csc z dz = – ctg z + c sen z = ? z = arc sen ? ? ? 2 ? ? csc z dz – ?1dz = – ctg z – z + c tg z = x 4 – x 2 ctg z = 4 – x 2 x Reemplazando ? 4 – x 2 – ctg z – z + c = – ? ? ? x ? ? – arc sen? 2 ? + c ? 4 – x 2 dx x 2 ? = – ? ? 4 – x 2 x ? ? – arc sen ? x ? + c ? ? 2 z x 4 – x 2 24
? ? ? 1 ? 1 1 ? ? ? ? ? ? ? ? 2 x 1 2 z – + c Ejercicios 7.3 Pag 571 Leythold Problema # 3 dx x x 2 + 4 x 2 + a 2 ? x = a tg z x 2 + 4 = x 2 + 2 2 ? x = 2 tg z dx x x 2 + 4 = ? 2 sec 2 z dz 2 tg z (2 sec z ) x = 2 tg z x2 = 4 tg2 z 1 sec z dz 2 tg z dz 1 cos z 2 sen z cos z ? dz 2 sen z si x = 2 tg z ? dx = 2 sec 2 z dz x 2 + 4 = 4 tg 2 z + 4 x 2 + 4 = 4 ? tg 2 z + 1? x 2 + 4 = 4 ? sec 2 z ? x 2 + 4 = 2 sec z 1 2 ? csc z dz si x = 2 tg z ? tg z = x 2 Tabla de integrales csc z = x 2 + 4 x ctg z = ? csc z dz = ln csc z – ctg z + c 1 2 ? csc z dz = Ln csc z – ctg z + c 4 + x 2 x Reemplazando 1 2 Ln csc z – ctg z + c = 1 2 Ln x 2 + 4 2 x x 2 ? x dx x 2 + 4 = 1 2 Ln x 2 + 4 – 2 x + c 25
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