Question

integration of dx/(x^2+a^2)^3/2

Answer 1

X= a tany …………… (1)

So, dx= a sec^2 y dy

Placing these values in the problem,without integration,

asec^2y dy/ (a^2 tan^2y+ a^2)^(3/2)

= asec^2y dy/ [ a^2 {tan^2y + 1} ]^(3/2)

= asec^2y dy/( asecy)^3

= asec^2y dy/ a^3 sec^3y

= dy / a^2 secy

= 1/a^2 (dy/ secy )

= 1/a^2 ( cosy dy)

So finally ,

Integration of dx/[ (x^2 + a^2 )^(3/2) ]

= 1/a^2 . integration of (cosy . dy )

= 1/ a^2 sin y + c …………. (2)

Now from the equation (1),

x = a tany

Or , tany = x/a

Or, tan^2 y = x^2 / a^2

Or, sec^2y = 1 + (x^2 / a^2)

Or , cos^2y = a^2 /( x^2 + a^2)

Or, sin^2 y = 1 - {a^2/ (x^2 + a^2)}

Or , sin^2 y = x^2 / (x^2+ a^2)

Or, siny = x/ [(x^2+ a^2)^(1/2)] .

So from the equation (2) ,

Integration of dx / [ ( x^2+ a^2)^(3/2) ]

= x/ [a^2 . ( x^2+ a^2)^(1/2) ] + c

Hope this may help u