{"id":8924,"date":"2024-01-28T04:51:31","date_gmt":"2024-01-27T23:21:31","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8924"},"modified":"2024-01-28T04:51:36","modified_gmt":"2024-01-27T23:21:36","slug":"integration-of-root-tanx","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/integration-of-root-tanx\/","title":{"rendered":"Integration of root tanx"},"content":{"rendered":"\n

Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by<\/p>\n\n\n\n

$\\int \\sqrt {tan x} \\; dx = \\frac {1}{\\sqrt 2} \\tan^{-1} (\\frac {tan x -1}{\\sqrt {2tan x}}) + \\frac {1}{2\\sqrt 2} \\ln \\frac {(tan x + 1 -\\sqrt {2tan x})}{(tan x + 1 +\\sqrt {2tan x})} + C$<\/p>\n\n\n\n

Proof of integration of root tanx<\/h2>\n\n\n\n

Let $t^2 = \\tan x$
$2tdt =\\sec^2 x dx$
Now
$\\sec^2 x = 1 + \\tan^2 x$
Therefore
$2t dt = (1+ t^4) dx $
$dx = \\frac {2t dt}{1+ t^4}$<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

$= \\int \\frac {2t^2}{1+ t^4} \\; dt $<\/p>\n\n\n\n

$= \\int \\frac {t^2 +1 + t^2 -1 }{1+ t^4} \\; dt $<\/p>\n\n\n\n

$\\int [\\frac {t^2 + 1}{1+ t^4} + \\frac {t^2 – 1}{1+ t^4}] \\; dt $<\/p>\n\n\n\n

$= \\int \\frac {t^2 + 1}{1+ t^4} \\; dt + \\int \\frac {t^2 – 1}{1+ t^4} \\; dt$<\/p>\n\n\n\n

$= \\int \\frac {1 + 1\/t^2}{t^2+ 1\/t^2} \\; dt + \\int \\frac {1 – 1\/t^2}{t^2+ 1\/t^2} \\; dt$<\/p>\n\n\n\n

$= \\int \\frac {1 + 1\/t^2}{(t – 1\/t)^2 +2} \\; dt + \\int \\frac {1 – 1\/t^2}{(t +1\/t)^2 -2} \\; dt$<\/p>\n\n\n\n

Let $u =t – 1\/t$ then $du = (1+ 1\/t^2) dt$
Let $v =t + 1\/t$ then $du = (1- 1\/t^2) dt$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$= \\int \\frac {1}{u^2 +2} \\; du + \\int \\frac {1 }{v^2 -2} \\; dv$<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

Now
$\\int \\frac {1}{x^2 + a^2} dx = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$
Proof<\/strong>
Put $x =a tan \\theta$ then $dx= a sec^2 \\theta d\\theta$
Therefore
$\\int \\frac {1}{x^2 + a^2} dx$
$=\\int \\frac {asec^2 \\theta}{a^2 tan^2 \\theta + a^2} d\\theta$
$=\\frac {1}{a} \\int d\\theta= \\frac {1}{a} \\theta + C = \\frac {1}{a} \\tan ^{-1} (\\frac {x}{a}) + C$<\/p>\n\n\n\n

$\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$
Proof<\/strong>
$\\frac {1}{x^2 – a^2} =\\frac {1}{2a}[ \\frac {1}{x-a} – \\frac {1}{x+a}]$
So
$\\int \\frac {1}{x^2 – a^2} dx $
$=\\frac {1}{2a}[ \\int \\frac {1}{x-a} dx – \\int \\frac {1}{x+a}]$
$= \\frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$<\/p>\n\n\n\n

Therefore the above integral becomes<\/p>\n\n\n\n

\\[
=\\frac {1}{\\sqrt 2} tan^{-1} \\frac {u}{\\sqrt 2} + \\frac {1}{2\\sqrt 2} ln |\\frac {v-\\sqrt 2}{v+ \\sqrt }| + C
\\]<\/p>\n\n\n\n

Now Substituting back the u and v<\/p>\n\n\n\n

\\[
=\\frac {1}{\\sqrt 2} tan^{-1} \\frac {t^2 -1}{\\sqrt 2 t} + \\frac {1}{2\\sqrt 2} ln |\\frac {t^2 + 1-\\sqrt 2 t}{t^2+ 1 + \\sqrt t }| + C
\\]<\/p>\n\n\n\n

Now Substituting back the t<\/p>\n\n\n\n

$\\int \\sqrt {tan x} \\; dx = \\frac {1}{\\sqrt 2} \\tan^{-1} (\\frac {tan x -1}{\\sqrt {2tan x}}) + \\frac {1}{2\\sqrt 2} \\ln \\frac {(tan x + 1 -\\sqrt {2tan x})}{(tan x + 1 +\\sqrt {2tan x})} + C$<\/p>\n\n\n\n

Definite Integration of root tanx<\/h2>\n\n\n\n

Generally we calculate the definite integral by finding the indefinite integral and then finding the difference, but we can use definite integral rules to simplify that. Let see how<\/p>\n\n\n\n

Example 1<\/strong><\/p>\n\n\n\n

$I = \\int_{0}^{\\pi\/2} \\sqrt {tan x} \\; dx$ -(1)<\/p>\n\n\n\n

Now we know that<\/p>\n\n\n\n

$\\int_{0}^{a} f(x) \\; dx =\\int_{0}^{a} f(a-x) \\; dx $<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$I=\\int_{0}^{\\pi\/2} \\sqrt {tan (\\pi\/2 – x} \\; dx= \\int_{0}^{\\pi\/2} \\sqrt {cot x} \\; dx$ -(2)<\/p>\n\n\n\n

Adding (1) and (2)<\/p>\n\n\n\n

$2I= \\int_{0}^{\\pi\/2} (\\sqrt {tan x} + \\sqrt {cot x}) \\; dx$<\/p>\n\n\n\n

$2I = \\int_{0}^{\\pi\/2} \\frac {sin x + cos x}{\\sqrt {sinx cos x}} \\; dx$<\/p>\n\n\n\n

$2I = \\sqrt 2 \\int_{0}^{\\pi\/2} \\frac {sin x + cos x}{\\sqrt {2sinx cos x}} \\; dx$<\/p>\n\n\n\n

$2I = \\sqrt 2 \\int_{0}^{\\pi\/2} \\frac {sin x + cos x}{\\sqrt {1 -(sinx – cosx)^2}} \\; dx$<\/p>\n\n\n\n

Taking $u=sin x – cos x$
$du =( cos x + sin x) dx$<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

$2I = \\sqrt 2 \\int_{-1}^{1} \\frac {1}{\\sqrt {1-u^2}} \\;du $
or<\/p>\n\n\n\n

$2I = 2 \\sqrt 2 \\int_{0}^{1} \\frac {1}{\\sqrt {1-u^2}} \\;du $
$I = \\sqrt 2 \\int_{0}^{1} \\frac {1}{\\sqrt {1-u^2}} \\;du $<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

$\\int \\frac {1}{\\sqrt {a^2 – x^2}} dx =\u00a0 \\sin ^{-1} (\\frac {x}{a}) + C$<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

$I = \\sqrt 2 [ \\pi\/2 – 0]= \\frac {\\pi}{\\sqrt 2} $<\/p>\n\n\n\n

\n

Other Integration Related Articles<\/p>\n\n\n\n

Integration of modulus function<\/a><\/td>Integration of tan x<\/a><\/td><\/td><\/td><\/tr>
Integration of sec x<\/a><\/td>Integration of greatest integer function<\/a><\/td><\/td><\/td><\/tr>
Integration of trigonometric functions<\/a><\/td>Integration of cot x<\/a><\/td><\/td><\/td><\/tr>
Integration of cosec x<\/a><\/td>Integration of sinx<\/a><\/td><\/td><\/td><\/tr>
Integration of irrational functions<\/a><\/td>integration of root x<\/a><\/td><\/td><\/td><\/tr>
Integration of fractional part of x<\/a><\/td>Integration of cos square x<\/a><\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by $\\int \\sqrt {tan x} \\; dx = \\frac {1}{\\sqrt 2} \\tan^{-1} (\\frac {tan x -1}{\\sqrt {2tan x}}) + \\frac {1}{2\\sqrt 2} \\ln \\frac {(tan x + 1 -\\sqrt {2tan x})}{(tan x + […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[498],"tags":[],"class_list":["post-8924","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nIntegration of root tanx - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/integration-of-root-tanx\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Integration of root tanx - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by $int sqrt {tan x} ; dx = frac {1}{sqrt 2} tan^{-1} (frac {tan x -1}{sqrt {2tan x}}) + frac {1}{2sqrt 2} ln frac {(tan x + 1 -sqrt {2tan x})}{(tan x + […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/physicscatalyst.com\/article\/integration-of-root-tanx\/\" \/>\n<meta property=\"og:site_name\" content=\"physicscatalyst's Blog\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/PhysicsCatalyst\" \/>\n<meta property=\"article:author\" content=\"https:\/\/www.facebook.com\/PhysicsCatalyst\" \/>\n<meta property=\"article:published_time\" content=\"2024-01-27T23:21:31+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-01-27T23:21:36+00:00\" \/>\n<meta name=\"author\" content=\"physicscatalyst\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"physicscatalyst\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"3 minutes\" \/>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Integration of root tanx - physicscatalyst's Blog","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/physicscatalyst.com\/article\/integration-of-root-tanx\/","og_locale":"en_US","og_type":"article","og_title":"Integration of root tanx - physicscatalyst's Blog","og_description":"Integration of root tan x is little complex. 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This can be solve using integration by substitution method . The formula is given by $\\int \\sqrt {tan x} \\; dx = \\frac {1}{\\sqrt 2} \\tan^{-1} (\\frac {tan x -1}{\\sqrt {2tan x}}) + \\frac {1}{2\\sqrt 2} \\ln \\frac {(tan x + 1 -\\sqrt {2tan x})}{(tan x +…","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8924","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=8924"}],"version-history":[{"count":1,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8924\/revisions"}],"predecessor-version":[{"id":8925,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/8924\/revisions\/8925"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=8924"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=8924"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=8924"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}