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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.


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