How do you integrate #int 1/(x^4sqrt(x^2+3))# by trigonometric substitution?

Question

Ezekiel Alexander · Accepted Answer

#sqrt(x^2+3)/(9x)-(x^2+3)^(3/2)/(27x^3)+C# #int dx/(x^4*sqrt(x^2+3))# After using #x=sqrt3*tany# and #dx=sqrt3*(secy)^2*dy# transforms, I found #int (sqrt3*(secy)^2*dy)/((sqrt3*tany)^4*sqrt((sqrt3*tany)^2+3))# =#int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt((sqrt3*secy)^2))# =#int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt3*secy)# =#1/9int (secy*dy)/(tany)^4# =#1/9int (coty)^4*secy*dy# =#1/9int (cosy/siny)^4*(dy/cosy)# =#1/9int ((cosy)^3*dy)/(siny)^4# =#1/9int ((cosy)^2*cosy*dy)/(siny)^4# =#1/9int ((1-(siny)^2)*cosy*dy)/(siny)^4# After using #z=siny# and #dz=cosy*dy# transforms, it became #1/9int ((1-z^2)*dz)/z^4# =#1/9int z^(-4)*dz-1/9int z^(-2)*dz# =#1/9z^(-1)-1/27z^(-3)+C# =#1/9(siny)^(-1)-1/27(siny)^(-3)+C# =#1/9cscy-1/27(cscy)^3+C# After using #x=sqrt3*tany#, #tany=x/sqrt3#, #secy=sqrt(x^2+3)/sqrt3# and #cscy=secy/tany=sqrt(x^2+3)/x# inverse transforms, I found #sqrt(x^2+3)/(9x)-(x^2+3)^(3/2)/(27x^3)+C#

Adam Adkins · Answer

undefined To integrate $ \int \frac{1}{{x^4\sqrt{x^2+3}}} $ by trigonometric substitution, follow these steps:

1. Substitute $ x = \sqrt{3}\tan{\theta} $.
2. Calculate $ dx $ using $ x = \sqrt{3}\tan{\theta} $.
3. Express $ x^2 + 3 $ in terms of $ \theta $.
4. Substitute $ dx $ and the expression for $ x^2 + 3 $ into the integral.
5. Simplify the integral using trigonometric identities.
6. Integrate the simplified expression.
7. Substitute $ \theta $ back in terms of $ x $ to find the final result.

Evelyn Agard · Answer

undefined To integrate $ \int \frac{1}{x^4\sqrt{x^2+3}} $ by trigonometric substitution, we can use the substitution $ x = \sqrt{3}\tan(\theta) $. After substituting and simplifying, we can then use trigonometric identities to express the integral in terms of trigonometric functions. Finally, we evaluate the integral in terms of theta. The detailed steps are as follows:

1. Let $ x = \sqrt{3}\tan(\theta) $. Then, $ dx = \sqrt{3}\sec^2(\theta) \, d\theta $.

2. Substitute $ x $ and $ dx $ into the integral:
$$ \int \frac{1}{(\sqrt{3}\tan(\theta))^4\sqrt{(\sqrt{3}\tan(\theta))^2+3}} \cdot \sqrt{3}\sec^2(\theta) \, d\theta $$

3. Simplify the expression inside the integral:
$$ = \int \frac{\sqrt{3}\sec^2(\theta)}{3\tan^4(\theta)\sqrt{3\tan^2(\theta)+3}} \, d\theta $$

4. Simplify further:
$$ = \int \frac{\sec^2(\theta)}{3\tan^4(\theta)} \, d\theta $$

5. Rewrite in terms of sine and cosine using the identity $ \tan^2(\theta) + 1 = \sec^2(\theta) $:
$$ = \int \frac{1}{3(\tan^2(\theta)+1)\tan^4(\theta)} \, d\theta $$

6. Rewrite $ \tan^2(\theta) $ as $ \sec^2(\theta) - 1 $:
$$ = \int \frac{1}{3((\sec^2(\theta) - 1)+1)\tan^4(\theta)} \, d\theta $$

7. Simplify inside the integral:
$$ = \int \frac{1}{3(\sec^2(\theta)\tan^4(\theta))} \, d\theta $$

8. Rewrite $ \sec^2(\theta)\tan^4(\theta) $ as $ (\sec^2(\theta))^2 - \sec^2(\theta) $:
$$ = \int \frac{1}{3(\sec^4(\theta) - \sec^2(\theta))} \, d\theta $$

9. Let $ u = \sec(\theta) $ and $ du = \sec(\theta)\tan(\theta) \, d\theta $:
$$ = \frac{1}{3} \int \frac{1}{u^4 - u^2} \, du $$

10. Factor the denominator:
$$ = \frac{1}{3} \int \frac{1}{u^2(u^2 - 1)} \, du $$

11. Perform partial fraction decomposition:
$$ = \frac{1}{3} \int \left(\frac{1}{u^2} - \frac{1}{u^2 - 1}\right) \, du $$

12. Integrate each term separately:
$$ = \frac{1}{3} \left(-\frac{1}{u} - \ln|u - 1| + \ln|u|\right) + C $$

13. Substitute back $ u = \sec(\theta) $ and simplify:
$$ = -\frac{1}{3\sec(\theta)} - \frac{1}{3}\ln|\sec(\theta) - 1| + \frac{1}{3}\ln|\sec(\theta)| + C $$

14. Finally, substitute back $ \theta $ using the original substitution $ x = \sqrt{3}\tan(\theta) $ to obtain the integral in terms of $ x $.

This process provides the integral of $ \frac{1}{x^4\sqrt{x^2+3}} $ by trigonometric substitution.