How do you integrate #int x/sqrt(3 + x^2)dx# using trigonometric substitution?

Question

Christopher Allers · Accepted Answer

#int x/sqrt(3+x^2)*dx=sqrt(x^2+3)+C# #int x/sqrt(3+x^2)*dx# After using #x=sqrt3tany# and #dx=sqrt3*(secy)^2*dy# transforms, this integral became #int sqrt3tany*(sqrt3*(secy)^2*dy)/(sqrt3*secy)# =#int sqrt3secy*tany*dy# =#sqrt3secy+C# After using #tany=x/sqrt3# and #secy=sqrt(x^2+3)/sqrt3# inverse transforms, I found #int x/sqrt(3+x^2)*dx=sqrt3*sqrt(x^2+3)/sqrt3+C=sqrt(x^2+3)+C#

Aurora Alvarado · Answer

undefined To integrate $ \frac{x}{\sqrt{3 + x^2}} $ using trigonometric substitution, we can let $ x = \sqrt{3} \tan(\theta) $.

Then $ dx = \sqrt{3} \sec^2(\theta) \, d\theta $.

Substituting these into the integral, we get:

$$ \int \frac{x}{\sqrt{3 + x^2}} \, dx = \int \frac{\sqrt{3} \tan(\theta)}{\sqrt{3 + 3\tan^2(\theta)}} \cdot \sqrt{3} \sec^2(\theta) \, d\theta $$

$$ = \int \frac{\sqrt{3} \tan(\theta)}{\sqrt{3(1 + \tan^2(\theta))}} \cdot \sqrt{3} \sec^2(\theta) \, d\theta $$

$$ = \int \frac{\sqrt{3} \tan(\theta)}{\sqrt{3 \sec^2(\theta)}} \cdot \sqrt{3} \sec^2(\theta) \, d\theta $$

$$ = \int \tan(\theta) \, d\theta $$

$$ = -\ln|\cos(\theta)| + C $$

Substitute back $ x = \sqrt{3} \tan(\theta) $, then $ \cos(\theta) = \frac{1}{\sqrt{1 + \tan^2(\theta)}} $.

$$ = -\ln\left|\frac{1}{\sqrt{1 + \tan^2(\theta)}}\right| + C $$

$$ = -\ln\left|\frac{1}{\sqrt{1 + \left(\frac{x}{\sqrt{3}}\right)^2}}\right| + C $$

$$ = -\ln\left|\frac{1}{\sqrt{1 + \frac{x^2}{3}}}\right| + C $$

$$ = \ln\left|\sqrt{1 + \frac{x^2}{3}}\right| + C $$

$$ = \ln\left(\sqrt{\frac{3 + x^2}{3}}\right) + C $$

$$ = \frac{1}{2} \ln(3 + x^2) - \frac{1}{2} \ln(3) + C $$