Do you actually need to carry out the integration?
If not, you can simply use the fact that the area of an ellipse is . And since you are only looking for the area above the x-axis, you just cut that in half to get:
Hello, varunnayudu!
This is a classic problem -- the area of an ellipse (well, semi-ellipse).
Find the area of the region bounded by and above the x-axis.
The graph is an ellipse, center at (0,0), major axis (ħa,0), minor axis (0, ħb)
Due its symmetry, we can find the area in Quadrant 1 and double.
Solving for
The area of the semi-ellipse is: .
Let:
Substitute: .
. . . . . .
Back-substitute: .
. . And we have: .
Evaluate: .
Perhaps my post was not clear.
Soroban's post explains this step clearly. Let: . Now substitute and in the integral. Since x is varying from -a to a, will vary from to . Substitute the limit also.
Now do you see what i have done?
This idea is called integration by substitution. Haven't you seen it before?