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 Calculus Calculus Math Forum

 March 11th, 2017, 05:14 PM #1 Newbie   Joined: Mar 2017 From: canada Posts: 2 Thanks: 0 Integrals in terms of I Hi guys, can you help me with the following: I=Integral (upper bound: pi/24 lower bound: 0) tan^10(4x)sec(4x)dx Express the value of: Integral (upper bound: pi/24 lower bound: 0) tan^12(4x)sec(4x)dx in terms of I. March 11th, 2017, 07:51 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra We have $$I = \int_0^{\frac{\pi}{24}} \tan^{10}{(4x)}\sec{(4x)}\,\mathrm dx$$ So \begin{align*} \int_0^{\frac{\pi}{24}} \tan^{12}{(4x)}\sec{(4x)}\,\mathrm dx &= \int_0^{\frac{\pi}{24}} \big(\sec^2{(4x)}-1\big)\tan^{10}{(4x)}\sec{(4x)}\,\mathrm dx \\ &= \int_0^{\frac{\pi}{24}} \underbrace{\tan^{10}{(4x)}\sec^2{(4x)}}_{\mathrm dv}\underbrace{\sec{(4x)}}_u\,\mathrm dx - I &\text{evaluate by parts} \\ &= \left. \frac{1}{44}\tan^{11}{(4x)}\sec{(4x)} \right|_0^{\frac{\pi}{24}} - \frac{1}{11} \int_0^{\frac{\pi}{24}} \tan^{12}{(4x)}\sec{(4x)}\,\mathrm dx - I \\ \frac{12}{11}\int_0^{\frac{\pi}{24}} \tan^{12}{(4x)}\sec{(4x)}\,\mathrm dx &= \frac{1}{22\sqrt{3^{11}}} - I \end{align*} I might have made an error or two in there due to working in my head, but the method is OK. Thanks from Country Boy Tags integrals, terms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chikis Algebra 5 July 28th, 2014 09:52 PM Jhenrique Calculus 0 May 15th, 2014 03:03 AM Agata78 Number Theory 5 January 14th, 2013 06:15 AM riponks Elementary Math 5 September 24th, 2009 03:50 AM Soha Algebra 4 December 19th, 2006 04:23 PM

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