Trig intergral help

**x5pyd3rx** · April 19th 2010, 02:32 PM

$integrate: \frac{(sec2x)^2}{(tan2x)^5} dx$

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

$\frac{1}{2} integrate<img src=$ u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\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 $6\frac{u^{0}}{0}$

**e^(i*pi)** · April 19th 2010, 02:35 PM

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

$\frac{1}{2} integrate<img src=$ u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\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:

$\int x^{-1} = \ln |x| + C$

**x5pyd3rx** · April 19th 2010, 04:07 PM

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

**harish21** · April 19th 2010, 05:38 PM

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

$\frac{1}{2} integrate<img src=$ u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\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 $6\frac{u^{0}}{0}$

How about this:

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

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

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

now integrate

$\int cot^{3}(2x) \times csc^{2} 2x dx$ using substution rule..

**Homeomorphism** · April 19th 2010, 05:47 PM

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.

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