My Math Forum (http://mymathforum.com/math-forums.php)
-   Calculus (http://mymathforum.com/calculus/)
-   -   Integrals in terms of I (http://mymathforum.com/calculus/339476-integrals-terms-i.html)

 jackhammer March 11th, 2017 05:14 PM

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.

 v8archie March 11th, 2017 07:51 PM

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.

 All times are GMT -8. The time now is 10:33 PM.

Copyright © 2019 My Math Forum. All rights reserved.