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: 7,268 Thanks: 2434 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. 

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