Make a change of coordinates:


The Jacobian for this transformation is

and has a determinant of

Note that we need to use the Jacobian in the other direction; that is, we've computed

but we need the Jacobian determinant for the reverse transformation (from

to

. To do this, notice that

we need to take the reciprocal of the Jacobian above.
The integral then changes to

Answer:
The function is decreasing in the following intervals
A. (0, 1)
C. (2, pi)
Step-by-step explanation:
To answer this question, imagine that you draw lines of slope m parallel to the function shown at each point.
-If the slope of this line parallel to the function is negative for those points then the function is decreasing.
-If the slope of this line parallel to the function is positive for those points then the function is increasing.
Observe in the lines drawn in the attached image. You can see that they have slope less than zero in the following interval:
(0, 1) U (2, pi)
Therefore the correct option is:
A. (0, 1)
C. (2, pi)
D ) x = 0 and x = 3 π / 4
.............................................
I'm pretty sure the answer is 12.