4x2 dxSolutionlet us substitute x 8sin dx 8cosd substitut

Solution

let us substitute x = 8sin

=> dx = 8cosd

substituting x=8sin and dx=8cosd we get

=>integral of 4*64*sin^2/((64(1-sin^2))

=>integral of 4*64*sin^2*8cosd/8cos

=>integral of 256sin^d

=>integral of 256(1-cos2)/2 *d

=>integral of 128(1-cos2)*d

=>128 - 128sin2/2

=>128 - 64 sin2

we have x=8sin

=>sin2 = 2sincos = 2*(x/8)*((64-x^2)/8)

=>= sin^-1(x/8)

=>128 - 64 sin2

=> 128sin^-1(x/8) - 64*(x/8)*((64-x^2)/4

=>128sin^-1(x/8) - 2*(x)*((64-x^2)) is the answer

i have explained very clearly please rate A+