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