Question

How do you integrate `int dx/sqrt(16+x^2)^2` by trigonometric substitution?

Answer 1

`" "`

Assuming you meant: `intdx/sqrt(16+x^2)`

Let `x=4tantheta`. This implies that `dx=4sec^2thetacolor(white).d theta`. Substituting these in yields:

`=int(4sec^2thetacolor(white).d theta)/sqrt(16+16tan^2theta)=4/sqrt16intsec^2theta/sqrt(1+tan^2theta)d theta`

Recall that `1+tan^2theta=sec^2theta`:

`=intsec^2theta/secthetad theta=intsecthetacolor(white).d theta=lnabs(tantheta+sectheta)+C`

Rewriting in terms of tangent since our substitution is `tantheta=x/4`:

`=lnabs(tantheta+sqrt(1+tan^2theta))+C=lnabs(x/4+sqrt(1+x^2/16))+C`

`=lnabs(x/4+1/4sqrt(16+x^2))+C`

Factoring the `1/4` and moving it out of the integral as the constant `ln(1/4)` through the `log(AB)=log(A)+log(B)` rule, we see it combines with the constant of integration:

`=lnabs(x+sqrt(16+x^2))+C`

`" "`

Assuming you meant: `intdx/(sqrt(16+x^2))^2`

The square root and the exponent cancel, leaving just:

`=intdx/(16+x^2)`

Now use the same substitution as before, `x=4tantheta` such that `dx=4sec^2thetacolor(white).d theta`.

`=int(4sec^2thetacolor(white).d theta)/(16+16tan^2theta)=1/4intsec^2theta/(1+tan^2theta)d theta=1/4intd theta`

`=1/4theta+C`

From the substitution `x=4tantheta` solving for `theta` yields:

`=1/4arctan(x/4)+C`