How do you integrate #int x^3/sqrt(64+x^2)# by trigonometric substitution?

4 Answers

#I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C#

Explanation:

Use the substitution #x= 8tantheta#. Then #dx= 8sec^2theta d theta#.

#I = int (8tantheta)^3/sqrt(64 + (8tantheta)^2) * 8sec^2theta d theta#

#I = int (512tan^3theta)/sqrt(64 + 64tan^2theta) * 8sec^2theta d theta#

#I = int(512tan^3theta)/sqrt(64(1 + tan^2theta)) * 8sec^2theta d theta#

Now use #1 + tan^2theta = sec^2theta#.

#I = int(512tan^3theta)/sqrt(64sec^2theta) * 8sec^2theta d theta#

#I = int(512tan^3theta)/(8sectheta) * 8sec^2theta d theta#

#I = int512sec thetatan^3theta d theta#

#I =int512secthetatantheta(tan^2theta) d theta#

We now use #tan^2theta = sec^2theta -1#.

#I = int512secthetatantheta(sec^2theta - 1) d theta#

Now let #u = sectheta#. Then #du = secthetatantheta d theta# and #d theta = (du)/(secthetatantheta)#.

#I = int512secthetatantheta * (u^2- 1) (du)/(secthetatantheta)#

#I= int(512u^2 - 512)du#

#I = 512/3u^3 - 512u + C#

#I = 512/3sec^3theta - 512sectheta#

From our initial substitution, we know that #tantheta = x/8#. Then #sectheta = sqrt(x^2 + 64)/8#

#I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C#

Hopefully this helps!

#int (x^3)/(sqrt{64+x^2}) "d"x= 1/3sqrt{64+x^2}(x^2-128) + C#.

Explanation:

I know you specified trigonometric substitution but I don't see why you'd use on in this case as a more obvious one stands out to me, because we have an #x^2# with an #x^3# on the numerator.

Someone's already submitted the trigonometric substitution method so thought I may as well share.

#I = int (x^3)/(sqrt{64+x^2})"d"x#.

Let #u=x^2#, #("d"u)/("d"x) = 2x#, #"d"x = 1/(2x) "d"u#.

Then,

#I = 1/2 int x^2/(sqrt{64+x^2}) "d"u#,

#= 1/2 int u(64+u)^(-1/2) "d"u#.

At this point, I used integration by parts. Let #s = u#, and #dt = (64 + u)^(-1/2)du#. Then, for #int sdt = st - int tds#:

#I = 1/2 ( [2u(64+u)^(1/2)]-2int(64+u)^(1/2) "d"u)#

#I = u(64+u)^(1/2) - 2/3(64+u)^(3/2) + C#

Then, I factored out #1/3 sqrt(64 + x^2)# and substituted #u = x^2# back in:

#I = 1/3sqrt{64+x^2}(3x^2-2(64+x^2)) + C#

#= color(blue)(1/3sqrt{64+x^2}(x^2-128) + C)#.

Jun 27, 2017

# 1/3sqrt(x^2+64)(x^2-128)+C.#

Explanation:

Here is another way to solve the Problem, without using

substitution.

#I=intx^3/sqrt(x^2+64)dx,#

#=int(x^2*x)/sqrt(x^2+64)dx,#

#=int{((x^2+64)-64)x}/sqrt(x^2+64)dx,#

#=int((x^2+64)/sqrt(x^2+64)-64/sqrt(x^2+64))xdx,#

#=int{(x^2+64)^(1/2)-64(x^2+64)^(-1/2)}{1/2d/dx(x^2+64)}dx,#

#=int(x^2+64)^(1/2){1/2d/dx(x^2+64)}dx#
#-64int(x^2+64)^(-1/2){1/2d/dx(x^2+64)}dx,#

#=1/2int(x^2+64)^(1/2)(d/dx(x^2+64))dx#
#-32int(x^2+64)^(-1/2)(d/dx(x^2+64))dx,#

#=1/2(x^2+64)^(1/2+1)/(1/2+1)-32(x^2+64)^(-1/2+1)/(-1/2+1),#

#=1/3(x^2+64)^(3/2)-64(x^2+64)^(1/2),#

#=1/3sqrt(x^2+64){(x^2+64)-192)},#

#=1/3sqrt(x^2+64)(x^2-128)+C.#

N.B.: The Final Integrals were obtained using the Rule,

#int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c, n ne-1.#

Enjoy Maths.!

Jul 9, 2017

After using #x=8sinhu# and #dx=8coshu# transforms, this integral became,

#int(8sinhu)^3*(8coshu)/(8coshu)*du#

#=int512(sinhu)^3*du#

#=int512(sinhu)^2*sinhu*du#

#=int512((coshu)^2-1)*sinhu*du#

#=int512(coshu)^2*sinhu*du-int512sinhu*du#

#=512/3*(coshu)^3-512coshu+C#

After using #sinhu=x/8# and #coshu=(Sqrt(x^2+64))/8# inverse transforms, solution of this integral became,

=#int(x^3)/(Sqrt(x^2+64))*dx#

=#=1/3*(x^2+64)^(3/2)-64Sqrt(x^2+64)+C#

=#=1/3*(x^2-128)*Sqrt(x^2+64)+C#