# [SOLVED] Definite Integral, parts, with trig substution

• Sep 8th 2008, 11:28 AM
matt3D
[SOLVED] Definite Integral, parts, with trig substution
Okay I have: $\int \frac {sec^{14}{20x}}{cot(20x)}$
I know I should break up $sec{x}$ but I am not sure what to do for the trig substitutions.
Thanks,
Matt
• Sep 8th 2008, 11:33 AM
Chris L T521
Quote:

Originally Posted by matt3D
Okay I have: $\int \frac {sec^{14}{20x}}{cot(20x)}$
I know I should break up $sec{x}$ but I am not sure what to do for the trig substitutions.
Thanks,
Matt

You're correct, you first need to split off a factor of $\sec(20x)$

$\int\frac{\sec^{14}(20x)}{\cot(20x)}\,dx=\int\frac {\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)}\,dx$

Now note that $\frac{1}{\cot(20x)}=\tan(20x)$

So we now see that $\int\frac{\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)} \,dx=\int \sec^{13}(20x)\cdot\sec(20x)\tan(20x)\,dx$

Now it should be clear what the proper substitution should be.

I hope this makes sense! (Sun)

--Chris
• Sep 9th 2008, 12:20 AM
matt3D
Thanks Chris! I forgot to mention that it had limits from $0$ to $pi/60$ after I did u substitution I got $\frac {16383}{280}$
• Sep 9th 2008, 03:13 AM
CaptainBlack
Quote:

Originally Posted by matt3D
Okay I have: $\int \frac {sec^{14}{20x}}{cot(20x)}$
I know I should break up $sec{x}$ but I am not sure what to do for the trig substitutions.
Thanks,
Matt

$\int_0^{\pi/60} \frac{(\sec(20x))^{14}}{\cot(20x)} \ dx= \int_0^{\pi/60} \frac{\sin(20x)}{(\cos(20x))^{15}} \ dx=$ $\int_0^{\pi/60} \frac{1}{20 \times 14}\frac{d}{dx} (\cos(20x))^{-14} \ dx$

RonL