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posted by  mrgrunchy on 5/30/2008 1:40:02 AM  |  status: Closed  

Please, Help needed very soon..Integrate sec^2(x)/tan^5(x)

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Calculus N/A N/A N/A
Question Details:
I am in need of assistance with the one problem i am having currently:
                                 
 
Dont know were to start...I am certain though its not as hard as it looks...just sick of it
please help..
Thanks,
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posted by chibba on 5/30/2008 1:50:37 AM  |  status: Live
Asker's Rating: Lifesaver   
mrgrunchy's comment:
"Thinking to hard..Thankyou"
Response Details:
 
let u = tanx                take the derivative
 
du = sec2x dx             divide sec2x to both side
 
du/sec2x = dx              plug in du/sec2x for dx
 
               simplify
 
                 integrate
 
                     plug in tanx for u and rewrite then plus C
 
Do not use L'Hopital's Rule on my post unless it's neccessary!
 
Feel free to msg me if you have any questions with my responses!
 
The indeterminate forms include 00, 0/0, 1∞, ∞ - ∞, ∞/∞, 0×∞, and ∞0.
Tags: Math, Calculus
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posted by gongalla on 5/30/2008 1:53:39 AM  |  status: Live
Asker's Rating: Helpful   
mrgrunchy's comment:
"Thank you, my professor would take the simpler of the two, but knowing different ways to solve the same problem is very helpful indeed...Thank you"
Response Details:


                   
                                   put cotx = t
                                         
                 
                   
 Hope this will help U
Tags: Math, Calculus
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posted by TheGreatPostulio on 5/30/2008 1:57:25 AM  |  status: Live
Asker's Rating: Somewhat Helpful   
Response Details:
secx = 1/cosx
and tanx = sinx/cosx
so sec^2x/tan^5x = (1/cos^2x)/(sin^5x/cos^5x) = cos^3x/sin^5x = (cotx)^3*(cscx)^2

take u = cotx
so du = -(cscx)^2 dx
so -du = cscx^2 dx
so int cot^3x csc^2x dx = int u^3(-du) = - u^4/4 + c
= -(cotx)^4/4 + c
Tags: Math, Calculus
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