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

## Expert Answers

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

so,

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)

so,

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

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