Answer:
y
=
m
x
+
b
in order to solve for
b
. Then, plug in known values.
y
=
0.6
x
+
0.6
Step-by-step explanation:
is this the answer u were looking for?
![\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B3x%7D%7B%5Ccos%5E2%282x%5E2%29%7D%5C%2C%5Cmathrm%20dx)
Substitute
, so that
:
![\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{4x}{\cos^2(2x^2)}\,\mathrm dx=\frac34\int\frac{\mathrm dy}{\cos^2y}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B3x%7D%7B%5Ccos%5E2%282x%5E2%29%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac34%5Cint%5Cfrac%7B4x%7D%7B%5Ccos%5E2%282x%5E2%29%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac34%5Cint%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Ccos%5E2y%7D)
Then
![\dfrac1{\cos^2y}=\sec^2y=\dfrac{\mathrm d}{\mathrm dy}[\tan y]](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Ccos%5E2y%7D%3D%5Csec%5E2y%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dy%7D%5B%5Ctan%20y%5D)
so that the integral wrt
comes out to be
![\displaystyle\frac34\int\sec^2y\,\mathrm dy=\frac34\tan y+C](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac34%5Cint%5Csec%5E2y%5C%2C%5Cmathrm%20dy%3D%5Cfrac34%5Ctan%20y%2BC)
Replace
to solve for the integral wrt
:
![\displaystyle\int\frac{3x}{\cos^2(2x^2)}\,\mathrm dx=\boxed{\frac34\tan(2x^2)+C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B3x%7D%7B%5Ccos%5E2%282x%5E2%29%7D%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B%5Cfrac34%5Ctan%282x%5E2%29%2BC%7D)
yo sup??
as mentioned earlier XUY is simply the summation of all the unique elements of X and Y
S=XUY
={-1,0,1,2,5}
Therefore your answer is option 1 ie
1.{-1,0,1,2,5}
Hope this helps.