Given:
The expression is:
To find:
The integration of the given expression.
Solution:
We need to find the integration of .
Let us consider,
![[\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)
![\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)
Therefore, the integration of is .
He worked as a carpenter for 12 hrs. and as a blacksmith for 18.
Answer:
2x+4
Step-by-step explanation:
A is the correct choose.could you make me brainlets answer please
From point A, draw a line segment at an angle to the given line, and about the same length. The exact length is not important. Set the compasses on A, and set its width to a bit less than one fifth of the length of the new line. Step the compasses along the line, marking off 5 arcs. Label the last one C. With the compasses' width set to CB, draw an arc from A just below it. With the compasses' width set to AC, draw an arc from B crossing the one drawn in step 4. This intersection is point D. Draw a line from D to B. Using the same compasses' width as used to step along AC, step the compasses from D along DB making 4 new arcs across the line. Draw lines between the corresponding points along AC and DB. Done. The lines divide the given line segment AB in to 5 congruent parts.
