`int sqrt(4+x^2) dx` Find the indefinite integral

Expert Answers

An illustration of the letter 'A' in a speech bubbles


Given to solve,

`int sqrt(4+x^2) dx`

using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows


for `sqrt(a+bx^2) ` we can take `x= sqrt(a/b) tan(u)`

so ,For

`int sqrt(4+x^2) dx`

the` x= sqrt(4/1)tan(u)= 2tan(u)`

=> `dx= 2sec^(2) (u) du`


`int sqrt(4+x^2) dx`

= `int sqrt(4+(2tan(u))^2) (2sec^(2) (u) du)`

= `int sqrt(4+4(tan(u))^2) (2sec^(2) (u) du)`

=`int sqrt(4(1+(tan(u))^2)) (2sec^(2) (u) du)`

= `int 2sqrt(1+tan^2(u))(2sec^(2) (u) du)`

= `int 2sec(u)(2sec^(2) (u) du)`

= `int 4sec^(3) (u) du`

`= 4int sec^(3) (u) du`

by applying the Integral Reduction

`int sec^(n) (x) dx`

`= (sec^(n-1) (x) sin(x))/(n-1) + ((n-2)/(n-1)) int sec^(n-2) (x) dx`

so ,

`4int sec^(3) (u) du`

= `4[(sec^(3-1) (u) sin(u))/(3-1) + ((3-2)/(3-1)) int sec^(3-2) (u)du]`

= `4[(sec^(2) (u) sin(u))/(2) + ((1)/(2)) int sec (u)du]`

=`4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

but` x= 2tan(u)`

=>` x/2 = tan(u)`

`u = tan^(-1) (x/2)`


`4[(sec^(2) (u) sin(u))/(2) + (1/2) (ln(tan(u)+sec(u)))]`

`=4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln(tan( tan^(-1) (x/2))+sec( tan^(-1) (x/2))))]`

=`4[(sec^(2) ( tan^(-1) (x/2)) sin( tan^(-1) (x/2)))/(2) + (1/2) (ln((x/2))+sec( tan^(-1) (x/2)))] +c`





See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial