Find the indefinite integral. integral csc^2(x\/5)dx Observe that is the derivative of 1\/5x. Let g(x) = 1\/5x. Multiply and divide the given integral integral csc^2(x\/5) dx by 1\/5. Therefore, the integral becomes integral csc^2 (x\/5)(1\/5) dx. Now, g(x) = x\/5, g'(x) = 1\/5. Define f(g(x)) = csc^2(x\/5), such that f(x) = csc^2 Rewrite the given integral in terms of g(x), where F(g(x)) is the antiderivative of f(g(x)). integral csc^2 (x\/5) dx = 5 integral csc^2 (x\/5)(1\/5)dx = 5 integral f((x))g'(x) dx = 5<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the indefinite integral of ( \\int \\csc^2\\left(\\frac{x}{5}\\right) \\, dx ), we will use substitution and the known properties of antiderivatives.<\/p>\n\n\n\n Notice that the integral involves ( \\csc^2\\left(\\frac{x}{5}\\right) ), which is a function of ( x ). We will simplify the integral using a substitution. Let:<\/p>\n\n\n\n [ Therefore, the derivative of ( g(x) ) is:<\/p>\n\n\n\n [ This substitution will help simplify the integral. Now, we can rewrite the integral in terms of ( g(x) ).<\/p>\n\n\n\n To apply the substitution correctly, we need to account for the factor of ( \\frac{1}{5} ). Thus, multiply and divide the given integral by ( \\frac{1}{5} ) to adjust for the substitution:<\/p>\n\n\n\n [ Now, the integral has a familiar form, where ( f(x) = \\csc^2(x) ) and the antiderivative of ( \\csc^2(x) ) is ( -\\cot(x) ). Therefore:<\/p>\n\n\n\n [ We now apply the antiderivative to the integral in terms of ( g(x) ):<\/p>\n\n\n\n [ Substitute ( g(x) = \\frac{x}{5} ) back into the equation:<\/p>\n\n\n\n [ The final result is:<\/p>\n\n\n\n [ In this solution, we used substitution to simplify the integral. By letting ( g(x) = \\frac{x}{5} ), we made the integrand easier to handle, turning it into a familiar form involving ( \\csc^2(x) ), whose antiderivative is well-known. This approach utilizes the chain rule, ensuring we multiply by the correct constant to account for the substitution.<\/p>\n\n\n\n In summary, substitution is a powerful technique that allows us to transform more complicated integrals into simpler ones by using known antiderivatives. In this case, recognizing the derivative relationship between ( \\csc^2(x) ) and ( \\cot(x) ) was crucial to finding the solution.<\/p>\n","protected":false},"excerpt":{"rendered":" Find the indefinite integral. integral csc^2(x\/5)dx Observe that is the derivative of 1\/5x. Let g(x) = 1\/5x. Multiply and divide the given integral integral csc^2(x\/5) dx by 1\/5. Therefore, the integral becomes integral csc^2 (x\/5)(1\/5) dx. Now, g(x) = x\/5, g'(x) = 1\/5. Define f(g(x)) = csc^2(x\/5), such that f(x) = csc^2 Rewrite the given […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"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":"","ast-disable-related-posts":"","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":"default","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":[25],"tags":[],"class_list":["post-191080","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191080","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=191080"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191080\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191080"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191080"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191080"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Set up the substitution<\/h3>\n\n\n\n
g(x) = \\frac{x}{5}
]<\/p>\n\n\n\n
g'(x) = \\frac{1}{5}
]<\/p>\n\n\n\nStep 2: Modify the integral<\/h3>\n\n\n\n
\\int \\csc^2\\left(\\frac{x}{5}\\right) \\, dx = 5 \\int \\csc^2\\left(g(x)\\right) g'(x) \\, dx
]<\/p>\n\n\n\n
\\int \\csc^2(x) \\, dx = -\\cot(x)
]<\/p>\n\n\n\nStep 3: Apply the antiderivative<\/h3>\n\n\n\n
5 \\int \\csc^2\\left(g(x)\\right) g'(x) \\, dx = 5 \\left[ -\\cot\\left(g(x)\\right) \\right]
]<\/p>\n\n\n\n
= 5 \\left[ -\\cot\\left(\\frac{x}{5}\\right) \\right] + C
]<\/p>\n\n\n\nStep 4: Final answer<\/h3>\n\n\n\n
\\int \\csc^2\\left(\\frac{x}{5}\\right) \\, dx = -5 \\cot\\left(\\frac{x}{5}\\right) + C
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n