Answer: i am posotive its 360 :D
Step-by-step explanation;
An expression is undefined when you divide by zero. A simple example is 1/0 which in english is "one over zero"
You cannot have zero in the denominator. Something like 0/1 is allowed as it is equal to 0. In other words, 0/1 = 0. But 1/0 is not allowed.
The original equation:
Integral of csc(5x)
Use a u sub:
u = 5x
du = 5dx
Simplify the du:
![\frac{1}{5} du = dx](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20du%20%3D%20dx)
Apply to the equation:
Integral csc(u)
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
du
Simplify:
Integral
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
![\frac{1}{sin(u)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bsin%28u%29%7D%20)
du
Factor out the constant:
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
Integral
![\frac{1}{sin(u)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bsin%28u%29%7D%20)
du
Use a second Substitution:
v = tan(
![\frac{u}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bu%7D%7B2%7D%20)
)
du =
![\frac{2}{1+v^{2} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B1%2Bv%5E%7B2%7D%20%7D%20)
dv
Applying to the equation:
![\frac{1}{5} Integral \frac{1}{ \frac{2v}{1+ v^{2} } } * \frac{2}{1+v^{2} } dv](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20Integral%20%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B2v%7D%7B1%2B%20v%5E%7B2%7D%20%7D%20%7D%20%2A%20%5Cfrac%7B2%7D%7B1%2Bv%5E%7B2%7D%20%7D%20dv)
Simplify:
![\frac{1}{5} Integral \frac{1}{v} dv](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20Integral%20%20%5Cfrac%7B1%7D%7Bv%7D%20dv)
Integrate:
![\frac{1}{5} ln(|v|)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20ln%28%7Cv%7C%29)
Insert back in your v and u:
v = tan(
![\frac{u}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bu%7D%7B2%7D%20)
)
u = 5x
This gives us the final equation (don't forget your constant):
![\frac{1}{5} ln(|tan( \frac{5x}{2} )|) + C](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20ln%28%7Ctan%28%20%5Cfrac%7B5x%7D%7B2%7D%20%29%7C%29%20%2B%20C)
(Thank you for making me write it out, I made a mistake on the original answer.)
Answer:
63/40, also known as 1 23/40.
Step-by-step explanation:
Uses = Subtract.
1. Find the LCD (Least Common Denominator) of 11/5 and 5/8. Can also be LCM (Least Common Multiple) of 5 and 8.
LCD = 40.
2. Make the denominators the same as the LCD.
![\frac{11*8}{5*8} - \frac{5*5}{8*5}](https://tex.z-dn.net/?f=%5Cfrac%7B11%2A8%7D%7B5%2A8%7D%20-%20%5Cfrac%7B5%2A5%7D%7B8%2A5%7D)
3. Simplify.
![\frac{88}{40} - \frac{25}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B88%7D%7B40%7D%20-%20%5Cfrac%7B25%7D%7B40%7D)
4. Simplify.
![\frac{63}{40}](https://tex.z-dn.net/?f=%5Cfrac%7B63%7D%7B40%7D)
Mixed Number = ![1\frac{23}{40}](https://tex.z-dn.net/?f=1%5Cfrac%7B23%7D%7B40%7D)