{"id":1957,"date":"2023-12-21T15:16:08","date_gmt":"2023-12-21T13:16:08","guid":{"rendered":"https:\/\/matematicastro.es\/?p=1957"},"modified":"2025-02-10T15:08:47","modified_gmt":"2025-02-10T13:08:47","slug":"integrales-indefinidas-calculo-de-primitivas-ii","status":"publish","type":"post","link":"https:\/\/matematicastro.es\/?p=1957","title":{"rendered":"Integrales indefinidas. C\u00e1lculo de primitivas (II)"},"content":{"rendered":"\n<p class=\"has-text-align-right has-text-color has-link-color wp-elements-306d65d1fefcbd4365b96f226b26da2a wp-block-paragraph\" style=\"color:#5e1e1e;font-size:16px\"><em>Todo lo que no se comprende, envenena<\/em><\/p>\n\n\n\n<p class=\"has-text-align-right has-text-color has-link-color wp-elements-e228c832a86f543ff69f2b0fcc5c1473 wp-block-paragraph\" style=\"color:#5e1e1e;font-size:16px\"><em><a href=\"https:\/\/es.wikipedia.org\/wiki\/Eugenio_d%27Ors\" data-type=\"link\" data-id=\"https:\/\/es.wikipedia.org\/wiki\/Eugenio_d%27Ors\" target=\"_blank\" rel=\"noreferrer noopener\">Eugenio D&#8217;ors<\/a><\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-text-color has-alpha-channel-opacity has-background\" style=\"background-color:#5e1e1e;color:#5e1e1e\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">En la <a href=\"https:\/\/matematicastro.es\/?p=263\" target=\"_blank\" rel=\"noopener\">entrada anterior sobre integrales indefinidas<\/a> se obtuvieron las siguientes:<\/p>\n\n\n\n<p class=\"has-text-align-center wp-block-paragraph\">\\[\\int{\\cos^2x\\,dx}=\\frac{x+\\text{sen}\\,x\\cos x}{2}+C\\]<\/p>\n\n\n\n<p class=\"has-text-align-center wp-block-paragraph\">\\[\\int{\\text{sen}^2x\\,dx}=\\frac{x-\\text{sen}\\,x\\cos x}{2}+C\\]<\/p>\n\n\n\n<p class=\"has-text-align-center wp-block-paragraph\">\\[\\int{x\\cos x\\,dx}=x\\,\\text{sen}\\,x+\\cos x+C\\]<\/p>\n\n\n\n<p class=\"has-text-align-center wp-block-paragraph\">\\[\\int{x\\,\\text{sen}\\,x\\,dx}=-x\\cos x+\\text{sen}\\,x+C\\]<\/p>\n\n\n\n<p class=\"has-text-align-center wp-block-paragraph\">\\[\\int{\\text{sen}\\,x\\cos x\\,dx}=\\frac{\\text{sen}^2x}{2}+C=-\\frac{\\cos^2x}{2}+C\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;Vamos a calcular un par de ellas m\u00e1s. Para ello utilizaremos algunas de las f\u00f3rmulas anteriores.<\/p>\n\n\n\n<div class=\"gkblock-5\" style=\"text-align: left;\">\n\t<p>\\[\\int{x\\,\\text{sen}^2x\\,dx}=\\begin{bmatrix}u=x&amp;\\text{;}&amp;du=dx\\\\dv=\\text{sen}^2x\\,dx&amp;\\text{;}&amp;v=\\frac{1}{2}(x-\\text{sen}\\,x\\cos x)\\end{bmatrix}=\\]<\/p>\n\t<p>\\[=\\frac{1}{2}x(x-\\text{sen}\\,x\\cos x)-\\frac{1}{2}\\int{(x-\\text{sen}\\,x\\cos x)\\,dx}=\\]<\/p>\n\t<p>\\[=\\frac{1}{2}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x-\\frac{1}{2}\\,\\frac{x^2}{2}+\\frac{1}{2}\\,\\frac{\\text{sen}^2x}{2}+C=\\]<\/p>\n\t<p>\\[=\\frac{1}{4}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x+\\frac{1}{4}\\text{sen}^2x+C\\]<\/p>\n<\/div>\n\n\n\n<div class=\"gkblock-5\" style=\"text-align: left;\">\n\t<p>\\[\\int{x\\cos^2x\\,dx}=\\int{x(1-\\text{sen}^2x)\\,dx}=\\int{x\\,dx}-\\int{x\\,\\text{sen}^2x\\,dx}=\\]<\/p>\n\t<p>\\[=\\frac{1}{2}x^2-\\left(\\frac{1}{4}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x+\\frac{1}{4}\\text{sen}^2x+C\\right)=\\]<\/p>\n\t<p>\\[=\\frac{1}{4}x^2+\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x-\\frac{1}{4}\\text{sen}^2x+C\\]<\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Si introduces la expresi\u00f3n <a href=\"http:\/\/www.wolframalpha.com\/input\/?i=x*(sin(x))%5E2\" target=\"_blank\" rel=\"noopener\">x*(sin(x))^2 en WolframAlpha<\/a> obtienes la integral indefinida:<\/p>\n\n\n\n<p class=\"has-text-align-left wp-block-paragraph\">\\[\\int{x\\,\\text{sen}^2x\\,dx}=\\frac{1}{8}\\left(2x(x-\\text{sen}\\,2x)-\\cos2x\\right)+C\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">que es equivalente a la obtenida anterioremente ya que<\/p>\n\n\n\n<p class=\"has-text-align-left wp-block-paragraph\">\\[\\frac{1}{8}\\left(2x(x-\\text{sen}\\,2x)-\\cos2x\\right)=\\frac{1}{8}(2x^2-2x\\,\\text{sen}\\,2x-\\cos2x)=\\]<\/p>\n\n\n\n<p class=\"has-text-align-left wp-block-paragraph\">\\[=\\frac{1}{4}x^2-\\frac{1}{4}x\\,2\\,\\text{sen}\\,x\\cos x-\\frac{1}{8}(\\cos^2x-\\text{sen}^2x)=\\]<\/p>\n\n\n\n<p class=\"has-text-align-left wp-block-paragraph\">\\[=\\frac{1}{4}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x-\\frac{1}{8}(1-2\\,\\text{sen}^2x)=\\]<\/p>\n\n\n\n<p class=\"has-text-align-left wp-block-paragraph\">\\[=\\frac{1}{4}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x+\\frac{1}{4}\\text{sen}^2x-\\frac{1}{8}\\]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Todo lo que no se comprende, envenena Eugenio D&#8217;ors En la entrada anterior sobre integrales indefinidas se obtuvieron las siguientes: \\[\\int{\\cos^2x\\,dx}=\\frac{x+\\text{sen}\\,x\\cos x}{2}+C\\] \\[\\int{\\text{sen}^2x\\,dx}=\\frac{x-\\text{sen}\\,x\\cos x}{2}+C\\] \\[\\int{x\\cos x\\,dx}=x\\,\\text{sen}\\,x+\\cos x+C\\] \\[\\int{x\\,\\text{sen}\\,x\\,dx}=-x\\cos x+\\text{sen}\\,x+C\\] \\[\\int{\\text{sen}\\,x\\cos x\\,dx}=\\frac{\\text{sen}^2x}{2}+C=-\\frac{\\cos^2x}{2}+C\\] &nbsp;Vamos a calcular un par de ellas m\u00e1s. Para ello utilizaremos algunas de las f\u00f3rmulas anteriores. \\[\\int{x\\,\\text{sen}^2x\\,dx}=\\begin{bmatrix}u=x&amp;\\text{;}&amp;du=dx\\\\dv=\\text{sen}^2x\\,dx&amp;\\text{;}&amp;v=\\frac{1}{2}(x-\\text{sen}\\,x\\cos x)\\end{bmatrix}=\\] \\[=\\frac{1}{2}x(x-\\text{sen}\\,x\\cos x)-\\frac{1}{2}\\int{(x-\\text{sen}\\,x\\cos x)\\,dx}=\\] \\[=\\frac{1}{2}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos x-\\frac{1}{2}\\,\\frac{x^2}{2}+\\frac{1}{2}\\,\\frac{\\text{sen}^2x}{2}+C=\\] \\[=\\frac{1}{4}x^2-\\frac{1}{2}x\\,\\text{sen}\\,x\\cos [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":1962,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5,4],"tags":[],"class_list":["post-1957","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-analisis-matematico","category-bachillerato","post-with-thumbnail","post-with-thumbnail-large"],"_links":{"self":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1957","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1957"}],"version-history":[{"count":6,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1957\/revisions"}],"predecessor-version":[{"id":2820,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/posts\/1957\/revisions\/2820"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=\/wp\/v2\/media\/1962"}],"wp:attachment":[{"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1957"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1957"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/matematicastro.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1957"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}