How do you integrate #int x^3sqrt(x^2-1)# using substitution?

Question

Charles Anctil · Accepted Answer

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

Let us take the subst. #x^2-1=t^2#, so, #x^2=t^2+1#, &, #2xdx=2tdt, or, xdx=tdt# Now, #I=intx^3sqrt(x^2-1)dx=intx^2sqrt(x^2-1)xdx# #=int(t^2+1)(sqrt(t^2))tdt=int(t^4+t^2)dt=t^5/5+t^3/3# #=t^3/15(3t^2+5)=t/15*t^2(3t^2+5)# #=sqrt(x^2-1)/15*(x^2-1){3(x^2-1)+5}# #=1/15(3x^2+2)(x^2-1)^(3/2)+C#.

Charles Anctil · Answer

Here is a third solution.

#intx^3sqrt(x^2-1) dx = int x^2sqrt(x^2-1) x dx# Let #u = x^2-1#, so that #du = 2xdx# and #x^2 = u+1#. The integral becomes #int (u+1)u^(1/2) 1/2 du = 1/2 int (u^(3/2)+u^(1/2))du# # = 1/2[2/5u^(5/2)+2/3u^(1/2)] +C# # = 1/15[3u^(5/2)+5u^(3/2)]+C# # = 1/15 u^(3/2)[3u+5]+C# Back-substituting gets us # = 1/15 (x^2-1)^(3/2)[3(x^2-1)+5]+C# # = 1/15 (x^2-1)^(3/2)(3x^2+2)+C#

Emma Aldridge · Answer

undefined To integrate $ \int x^3\sqrt{x^2-1} $ using substitution, let $ u = x^2 - 1 $. Then, $ du = 2x \, dx $. Solving for $ dx $, we have $ dx = \frac{du}{2x} $. Substituting $ u $ and $ dx $, the integral becomes $ \frac{1}{2}\int u^{\frac{3}{2}} \, du $. Integrating $ u^{\frac{3}{2}} $ with respect to $ u $ gives $ \frac{2}{5}u^{\frac{5}{2}} + C $, where $ C $ is the constant of integration. Finally, substituting back $ x^2 - 1 $ for $ u $, the result is $ \frac{2}{5}(x^2 - 1)^{\frac{5}{2}} + C $.