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 March 11th, 2017, 06:14 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, 08:51 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,857 Thanks: 2230 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

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