How do you find the integral of #int x^3 * sqrt(x^2 + 4) dx#?

Question

Aurora Alvarado · Accepted Answer

#1/15(x^2+4)^(3/2)*(3x^2-8)+C# #int x^3sqrt(x^2+4)*dx# After using #x=2tany# and #dx=2(secy)^2*dy# transforms, this integral became, #int (2tany)^3sqrt((2tany)^2+4)*2(secy)^2*dy# =#int 16(tany)^3*(secy)^2*sqrt(4(secy)^2)*dy# =#int 32(secy)^3*(tany)^3*dy# =#int 32(secy)^2*(tany)^2*secy*tany*dy# =#int 32(secy)^2*((secy)^2-1)*secy*tany*dy# After using #z=secy# and #dz=secy*tany*dy# transforms, it became #int 32z^2*(z^2-1)*dz# =#int (32z^4-32z^2)*dz# =#32/5z^5-32/3z^3+C# =#32/5(secy)^5-32/3(secy)^3+C# After using #x=2tany#, #tany=x/2# and #secy=sqrt(x^2+4)/2# inverse transforms, I found #1/5(x^2+4)^(5/2)-4/3(x^2+4)^(3/2)+C# =#1/15(x^2+4)^(3/2)*(3x^2-8)+C#

Ezra Adams · Answer

# 1/15(x^2+4)^(3/2)(3x^2-8)+C#.

Let, #I=intx^3sqrt(x^2+4)dx=intx^2sqrt(x^2+4)*xdx#. Subst. #x^2+4=t^2, or, x^2=t^2-4. :. 2xdx=2tdt, or, xdx=tdt# #:. I=int(t^2-4)*sqrt(t^2)*tdt=int(t^2-4)t^2dt#, #=int(t^4-4t^2)dt#, #=t^5/5-4*t^3/3#, #=t^3/15(3t^2-20)#, #=1/15*t^2(3t^2-20)t#. Reverting from #t^2 to (x^2+4)#, we have, #I=1/15(x^2+4){3(x^2+4)-20}sqrt(x^2+4)#. # rArr I=1/15(x^2+4)^(3/2)(3x^2-8)+C#, as Respected Cem Sentin has derived easily. Take Joy in Mathematics!

Evelyn Agard · Answer

undefined To find the integral of $ \int x^3 \sqrt{x^2 + 4} \, dx $, you can use trigonometric substitution. Let $ x = 2 \tan(\theta) $. Then, $ dx = 2 \sec^2(\theta) \, d\theta $. Substitute these expressions into the integral and simplify using trigonometric identities. After integrating, express the result back in terms of $ x $.