A proportion is a pair of equal ratios.
Given:
The expression is:
![\dfrac{1-\cos 2x}{1+\cos 2x}](https://tex.z-dn.net/?f=%5Cdfrac%7B1-%5Ccos%202x%7D%7B1%2B%5Ccos%202x%7D)
To find:
The integration of the given expression.
Solution:
We need to find the integration of
.
Let us consider,
![I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx](https://tex.z-dn.net/?f=I%3D%5Cint%20%5Cdfrac%7B1-%5Ccos%202x%7D%7B1%2B%5Ccos%202x%7Ddx)
![[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]](https://tex.z-dn.net/?f=%5B%5Cbecause%201%2B%5Ccos%202x%3D2%5Ccos%5E2x%2C1-%5Ccos%202x%3D2%5Csin%5E2x%5D)
![I=\int \dfrac{\sin^2x}{\cos^2x}dx](https://tex.z-dn.net/?f=I%3D%5Cint%20%5Cdfrac%7B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7Ddx)
![\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Ctan%20%5Ctheta%20%3D%5Cdfrac%7B%5Csin%20%5Ctheta%7D%7B%5Ccos%20%5Ctheta%7D%5Cright%5D)
It can be written as:
![[\because 1+\tan^2 \theta =\sec^2 \theta]](https://tex.z-dn.net/?f=%5B%5Cbecause%201%2B%5Ctan%5E2%20%5Ctheta%20%3D%5Csec%5E2%20%5Ctheta%5D)
![I=\int \sec^2xdx-\int 1dx](https://tex.z-dn.net/?f=I%3D%5Cint%20%5Csec%5E2xdx-%5Cint%201dx)
![I=\tan x-x+C](https://tex.z-dn.net/?f=I%3D%5Ctan%20x-x%2BC)
Therefore, the integration of
is
.
Answer:
4, 6, 8, 10, 30
Step-by-step explanation:
Because when you multiply by 10, it adds a zero to the end of the number, like 12*10 = 120, so vise versa for dividing