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.