Trouble with trig identities in solving integral.

[NewtonianAlch]

Homework Statement

[itex]\int[/itex][itex]\frac{sec^{2}x}{tan^{4}x}[/itex]dx

Homework Equations

The Attempt at a Solution

I have the answer as -1/3 cot^3(x) + C listed.

All the intermediate steps are given, but the first one is they have converted the sec/tan integral in to:

[itex]\int[/itex]{cot^{2}x}{csc^{2}x} dx

I am a little confused by the trig identities used and manipulated to get this.

 

[cragar]

you can just write it all in terms of sines and cosines and then reduce it to get that.
without using any identities

 

[TheoMcCloskey]

Hint: what's the derivative of [itex]tan(x)[/itex]?

 

[SammyS]

Homework Statement

[tex]\int\frac{\sec^{2}x}{\tan^{4}x} dx[/tex]
...

I am a little confused by the trig identities used and manipulated to get this.

[tex]\frac{\sec^{2}(x)}{\tan^{4}(x)}=\frac{1}{\cos^2(x)} \frac {\cos^4(x)}{\sin^4(x)}[/tex]

Try it from there.

 

[Mark44]

Following TheoMcCloskey's hint, the problem can be done without converting to sines and cosines. After a fairly simple substitution, the integral looks like
[tex]\int u^{-4}du[/tex]

 

[NewtonianAlch]

Thanks for the responses guys, yes the d/dx (tan x) substitution can be done to get a result of

-1/3*1/tan^3 x + C

But there's that other result which is bugging me, I will take a closer look at the post with the other identities as I want to see how I can get the cot^2csc^2 result.

 

[SammyS]

Taking it one step further ...

[itex]\displaystyle \frac{\sec^{2}(x)}{\tan^{4}(x)}=\frac{1}{\cos^2(x) } \frac {\cos^4(x)}{\sin^4(x)}[/itex]

[itex]\displaystyle =\frac{1}{\cos^2(x) }\frac {\cos^2(x)}{\sin^2(x)}\frac {\cos^2(x)}{\sin^2(x)}[/itex]​