The Integral of 5x*csc(7x)*cot(7x)

[NastyAccident]

Homework Statement

Integral of 5x*csc(7x)*cot(7x)

Homework Equations

Integration by parts
Trigonometric Derivatives (csc(u) = -csc(u)cot(u))
Substitution

The Attempt at a Solution

*Note, I apologize beforehand for the lack of knowledge on how to display mathematical equations nicely on PF *

5*int[x*csc(7x)*cot(7x)]

I am preparing for the 7x substitution here:
5/7*int[7x*csc(7x)*cot(7x)]

let z = 7x
dz = 7dx

5/7 * 1/7 *int[z*csc(z)*cot(z)]
5/49 * int[z*csc(z)*cot(z)]

Preparing for the csc(z) substitution here:
-1* 5/49 * int[-1*z*csc(z)*cot(z)]

let w = csc(z)
dw = -csc(z)cot(z)

arccsc(w) = z

-1*5/49*int[arccsc(w)]

Integration by parts

u = arccsc(w) dv=dw
du = -1/|w|*sqrt(w^2-1) v = w

-1*5/49*[w*arccsc(w) - int[ -1/|w|*sqrt(w^2-1) * w ] ]

-1*5/49*[w*arccsc(w) + int[w/|w|*sqrt(w^2-1)]

In between these two steps I may have made an error since I canceled the absolute value w on the denominator with the w on the numerator.

-1*5/49*[w*arccsc(w) + int[1/sqrt(w^2-1)]

Solved for answer in two different ways (both wrong)
arccosh way -
-1*5/49*[w*arccsc(w) + arccosh(w)]
-1*5/49*[csc(z)*arccsc(csc(z)) + arccosh(csc(z))]
-1*5/49*[csc(7x)*arccsc(csc(7x)) + arccosh(csc(7x))]

trig substitution way -
-1*5/49*[w*arccsc(w) + int[1/sqrt(w^2-1)]

let w = 1*sec(theta)
dw = sec(theta)tan(theta)

-1*5/49*[w*arccsc(w) + int[sec(theta)tan(theta)/sqrt(sec(theta)^2-1)]
-1*5/49*[w*arccsc(w) + int[sec(theta)tan(theta)/tan(theta)]
-1*5/49*[w*arccsc(w) + int[sec(theta)]
-1*5/49*[w*arccsc(w) + ln|sec(theta)+tan(theta)|]

Building the right triangle

Since w/1 = sec(theta)

w is the hyp, 1 is the adj side, & sqrt(w^2-1) is then the opposite side

w -> /| <- sqrt(w^2-1)
/ |
---
^ (1)

tan(theta) = sqrt(w^2-1)
sec(theta) = w

-1*5/49*[w*arccsc(w) + ln|w+sqrt(w^2-1)|]

Final answer:
-1*5/49*[csc(7x)*arccsc(csc(7x)) + ln|csc(7x)+sqrt(csc(7x)^2-1)|]


So, I'm unsure as to why both of these answers were marked wrong (I've submitted them into webassign). Any help would be appreciated!

http://img168.imageshack.us/img168/3822/mathwoesonwebassign5xcs.th.jpg

 

[gabbagabbahey]

Preparing for the csc(z) substitution here:
-1* 5/49 * int[-1*z*csc(z)*cot(z)]

let w = csc(z)
dw = -csc(z)cot(z)

arccsc(w) = z

This looks like a problem to me; trig functions are periodic, and assign the same value to more than one input (eg. sin(0)=sin(pi)=0 ), arccsc(w) will return the so-called principle value only, and there is no guarantee that that value is equal to 'z' (it could be z+2pi or z+4pi...etc.)

Instead, try applying integration by parts directly to \int -z\csc z\cot z dz by using u=z and dv=-\csc z\cot z dz

 

[NastyAccident]

So:

5*int[x*csc(7x)*cot(7x)]

5/7*int[7x*csc(7x)*cot(7x)]

let z = 7x
dz = 7dx

5/7 * 1/7 *int[z*csc(z)*cot(z)]
5/49 * int[z*csc(z)*cot(z)]

-5/49 * int[z*-csc(z)*cot(z)]

====== (Working the problem from your advice now) ======

u = z dv = -csc(z)cot(z)
du = dz v = csc(z)

-5/49 *(z*csc(z) - int[csc(z)])

-5/49 *(z*csc(z) - ln|csc(z) - cot(z)|)

Plugging in for z.

-5/49 *(7x*csc(7x) - ln|csc(7x)-cot(7x)|) + C

Should be my final answer... (Can anyone verify that please?)

 

[gabbagabbahey]

Looks good to me!

 

[Jay J]

Yup seems right :)

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