Integrate #intx^3/sqrt(x^2+4)# using trig substitution?

Question

Emily Ammerman · Accepted Answer

See the explanatiom below

You have to change as follows #I=8(1/3u^3-u)# #I=8/3(sec^3theta-3sectheta)# #=8/3(((x^2+1)/2)^(3/2)-3sec(arctan(x/2))+C# It's easier without trigonometric substitution Let #u=x^2+4#, #=>#, #du=2xdx# #I=1/2int((u-4)du)/sqrtu# #=1/2intsqrtudu-int4/sqrtudu# #=(u^(3/2)/3-4sqrtu)# #=1/3(x^2+4)^(3/2)-4sqrt(x^2+4)# #=((x^2-8))/3sqrt(x^2+4)+C#

Isabella Ackerman · Answer

undefined To integrate $\int \frac{x^3}{\sqrt{x^2 + 4}}$ using trigonometric substitution, perform the following steps:

1. Let $x = 2\tan(\theta)$.
2. Find $\frac{dx}{d\theta}$ and express $x^2$ in terms of $\theta$.
3. Substitute $x$ and $dx$ into the integral.
4. Express $x^3$ in terms of $\theta$.
5. Replace $x^3$ and $\sqrt{x^2 + 4}$ with expressions involving $\theta$.
6. Simplify the resulting integral in terms of $\theta$.
7. Integrate with respect to $\theta$.
8. Substitute back the value of $x$ in terms of $\theta$.
9. Simplify the final expression.

Here are the steps in detail:

1. Let $x = 2\tan(\theta)$.
2. $dx = 2\sec^2(\theta)d\theta$.
3. $x^2 = (2\tan(\theta))^2 = 4\tan^2(\theta) = 4(\sec^2(\theta) - 1) = 4\sec^2(\theta) - 4$.
4. $x^3 = (2\tan(\theta))^3 = 8\tan^3(\theta)$.
5. $\sqrt{x^2 + 4} = \sqrt{4\sec^2(\theta) - 4} = 2\sqrt{\sec^2(\theta) - 1} = 2\tan(\theta)$.
6. Substitute $x^3$ and $\sqrt{x^2 + 4}$ into the integral:
   $\int \frac{8\tan^3(\theta)}{2\tan(\theta)} \cdot 2\sec^2(\theta)d\theta$.
7. Simplify:
   $\int 4\tan^2(\theta)\sec^2(\theta)d\theta$.
8. Integrate with respect to $\theta$:
   $= \int 4(\sec^4(\theta) - 1)d\theta$.
9. $= 4\int \sec^4(\theta)d\theta - 4\int d\theta$.
10. The integral of $\sec^4(\theta)$ can be evaluated using integration by parts.
11. Let $u = \sec^2(\theta)$ and $dv = \sec^2(\theta)d\theta$.
12. Find $du$ and $v$, then apply the integration by parts formula.
13. After integration by parts, simplify the expression.
14. Integrate $4\int d\theta$.
15. Substitute back the value of $x$ in terms of $\theta$ and simplify the final expression.

This completes the integration of $\int \frac{x^3}{\sqrt{x^2 + 4}}$ using trigonometric substitution.