# Trig intergral help

• Apr 19th 2010, 03:32 PM
x5pyd3rx
Trig intergral help
$integrate: \frac{(sec2x)^2}{(tan2x)^5} dx$

After i convert to tangets and such and do the u substitution i end up with this

$\frac{1}{2} integrate:( u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3})$

I must have gone wrong somewhere becuase if you integrate $6u^{-1}$ you get $6\frac{u^{0}}{0}$
• Apr 19th 2010, 03:35 PM
e^(i*pi)
Quote:

Originally Posted by x5pyd3rx
$integrate: \frac{(sec2x)^2}{(tan2x)^5} dx$

After i convert to tangets and such and do the u substitution i end up with this

$\frac{1}{2} integrate:( u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3})$

I must have gone wrong somewhere becuase if you integrate $6u^{-1}$ you get $6\frac{u^{0}}{0}$

You don't use the power law on x^-1 because you'd get the issue you described. Integrating $x^{-1}$ is a special case:

$\int x^{-1} = \ln |x| + C$
• Apr 19th 2010, 05:07 PM
x5pyd3rx
so with those changes does this seem correct? $\frac{1}{-6u^3} + \frac{4}{-4u^2} + 3\ln{u} + 4\frac{u^5}{10} + \frac{u^4}{8} +c$
• Apr 19th 2010, 06:38 PM
harish21
Quote:

Originally Posted by x5pyd3rx
$integrate: \frac{(sec2x)^2}{(tan2x)^5} dx$

After i convert to tangets and such and do the u substitution i end up with this

$\frac{1}{2} integrate:( u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3})$

I must have gone wrong somewhere becuase if you integrate $6u^{-1}$ you get $6\frac{u^{0}}{0}$

$\frac{sec^{2}(2x)}{tan^{5}(2x)} = \frac{1}{cos^{2}(2x)} \times \frac{1}{tan^{3}(2x)} \times \frac{1}{tan^{2}(2x)}$

$= \frac{1}{cos^{2}(2x)} \times \frac{1}{tan^{3}(2x)} \times \frac{cos^{2}(2x)}{sin^{2}(2x)}$

$= cot^{3}(2x) \times csc^{2} 2x$

now integrate

$\int cot^{3}(2x) \times csc^{2} 2x dx$ using substution rule..
• Apr 19th 2010, 06:47 PM
Homeomorphism
Quote:

Originally Posted by x5pyd3rx
so with those changes does this seem correct? $\frac{1}{-6u^3} + \frac{4}{-4u^2} + 3\ln{u} + 4\frac{u^5}{10} + \frac{u^4}{8} +c$

Not sure what you have done here, but, for $\int\dfrac{\left(\sec{2x}\right)^2}{\left(\tan{2x} \right)^5}\;{dx}$, if you let $u = \tan{2x}$, then $\dfrac{du}{dx} = 2\sec^2{2x}$ $\Rightarrow dx = \dfrac{du}{2\left(\sec{2x}\right)^2}$; so we have $\int\dfrac{\left(\sec{2x}\right)^2}{\left(\tan{2x} \right)^5}\;{dx} = \int\dfrac{\left(\sec{2x}\right)^2}{2u^5\left(\sec {2x}\right)^2}\;{du}$, which should be easier to finish off.