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

Question

William Abdalla · Accepted Answer

#intx/sqrt(16+x^2)dx=ln|sqrt(x^2+16)+x|+C#

For integrals involving the root #sqrt(x^2+a^2),# we can use the substitution #x=atantheta#. Here, #a^2=16, a=4, x=4tantheta#

So, #dx=4sec^2thetad theta# and we can apply the substitution.

#int(4sec^2theta)/sqrt(16+16tan^2theta)d theta=int(4sec^2theta)/(sqrt(16(1+tan^2theta)))d theta#

#=intsec^2theta/sqrt(1+tan^2theta)d theta#

Recalling the identity #1+tan^2theta=sec^2theta,#

#=intsec^2theta/sqrt(sec^2theta)d theta#

#=intsec^cancel(2)theta/cancel(secthetad) theta#

#=intsecthetad theta=ln|sectheta+tantheta|+C#

This is a fairly common integral, it should be memorized. We need things in terms of #x.# Recalling that #x=4tantheta, tantheta=x/4.#

We still need the secant in terms of #x.# Applying the identity #1+tan^2theta=sec^2theta:#

#16/16+x^2/16=sec^2theta#

#sectheta=sqrt(x^2+16)/4#

Thus, we have

#ln|(sqrt(x^2+16)+x)/4|+C=ln|sqrt(x^2+16)+x|-ln(4)+C#

We may absorb the #ln4# into the constant of integration. We're left with

#intx/sqrt(16+x^2)dx=ln|sqrt(x^2+16)+x|+C#

William Abdalla · Answer

undefined To integrate $ \int \frac{x}{\sqrt{16 + x^2}} \, dx $ using trigonometric substitution, we can let $ x = 4 \tan(\theta) $. Then, $ dx = 4 \sec^2(\theta) \, d\theta $. Substituting these expressions into the integral gives us:

$$ \int \frac{4\tan(\theta)}{\sqrt{16 + (4\tan(\theta))^2}} \cdot 4\sec^2(\theta) \, d\theta $$

Simplifying inside the square root yields:

$$ \int \frac{4\tan(\theta)}{\sqrt{16 + 16\tan^2(\theta)}} \cdot 4\sec^2(\theta) \, d\theta $$
$$ = \int \frac{4\tan(\theta)}{\sqrt{16(1 + \tan^2(\theta))}} \cdot 4\sec^2(\theta) \, d\theta $$
$$ = \int \frac{4\tan(\theta)}{\sqrt{16\sec^2(\theta)}} \cdot 4\sec^2(\theta) \, d\theta $$
$$ = \int \frac{4\tan(\theta)}{4\sec(\theta)} \cdot 4\sec^2(\theta) \, d\theta $$
$$ = \int \tan(\theta) \cdot 4\sec(\theta) \, d\theta $$

Using the identity $ \sec(\theta) = \frac{1}{\cos(\theta)} $, we can rewrite $ \tan(\theta) $ as $ \sin(\theta) \cdot \cos(\theta) $, giving:

$$ = \int \frac{4\sin(\theta) \cdot \cos(\theta)}{\cos(\theta)} \, d\theta $$
$$ = \int 4\sin(\theta) \, d\theta $$
$$ = -4\cos(\theta) + C $$

Finally, we need to express the result in terms of $ x $. Using the original substitution $ x = 4 \tan(\theta) $, we can find $ \theta $ using $ \tan(\theta) = \frac{x}{4} $, and then use $ \sec(\theta) = \sqrt{1 + \tan^2(\theta)} $ to find $ \cos(\theta) $. Substituting these values into the integral result gives the final answer:

$$ \int \frac{x}{\sqrt{16 + x^2}} \, dx = -4\sqrt{16 + x^2} + C $$