Originally Posted by
ticbol Here is one way.
INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx
Let x = 3sinA ----------***
So,
dx = 3cosA dA ----**
The new boundaries, for dA,
When x = 0,
0 = 3sinA
0 = sinA
A = arcsin(0) = 0 ----**
When x = 3/sqrt(2),
3/sqrt(2) = 3sinA
1/sqrt(2) = sinA
A = arcsin(1/sqrt(2)) = pi/4 ---**
(9 -x^2)^(-3/2)
= (9 -9sin^2(A))^(-3/2)
= 1 / [9(1-sin^2(A)]^3/2
= 1 / 27*(cos^2(A))^(3/2)
= 1 / 27cos^3(A)
So,
INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx
= INT.(0-->pi/4)[1 / 27cos^3(A)]*3cosA dA
= (3/27)INT.(0-->pi/4)[1 / cos^3(A)]*cosA dA
= (1/9)INT.(0-->pi/4)[1 / cos^2(A)]dA
= (1/9)INT.(0-->pi/4)[sec^2(A)]dA
= (1/9)[tanA](0-->pi/4)
= (1/9)[tan(pi/4) -tan(0)]
= (1/9)[1 -0]
= 1/9 ---------------------answer.