MHB Solving an integral with trigonometric substitution

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SUMMARY

The integral $$\int_{}^{} \frac {x^2}{{(4 - x^2)}^{3/2}}\,dx$$ can be solved using the trigonometric substitution $x = 2\sin\theta$. This substitution leads to $dx = 2\cos\theta d\theta$ and simplifies the expression for $\sqrt{4 - x^2}$ to $2\cos\theta$. The identity $\sin^2\theta + \cos^2\theta = 1$ is crucial for deriving this relationship, confirming that $\sqrt{4 - x^2} = 2\cos\theta$ is valid.

PREREQUISITES
  • Understanding of trigonometric identities, specifically $\sin^2\theta + \cos^2\theta = 1$
  • Familiarity with integral calculus and substitution methods
  • Knowledge of trigonometric functions and their properties
  • Basic skills in manipulating algebraic expressions
NEXT STEPS
  • Study trigonometric substitution techniques in integral calculus
  • Learn how to apply the identity $\sin^2\theta + \cos^2\theta = 1$ in various contexts
  • Explore more complex integrals involving trigonometric functions
  • Practice solving integrals using different substitution methods
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, as well as educators teaching integral calculus and trigonometric substitutions.

tmt1
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I have this integral:

$$\int_{}^{} \frac {x^2}{{(4 - x^2)}^{3/2}}\,dx$$

I can see that we can substitute $x = 2sin\theta$, and $dx = 2cos\theta d\theta$, but I am unable to see how $\sqrt{4 - x^2} = 2cos\theta$. How can I get this substitution?
 
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tmt said:
I have this integral:

$$\int_{}^{} \frac {x^2}{{(4 - x^2)}^{3/2}}\,dx$$

I can see that we can substitute $x = 2sin\theta$, and $dx = 2cos\theta d\theta$, but I am unable to see how $\sqrt{4 - x^2} = 2cos\theta$. How can I get this substitution?
Recall the identity:. . \sin^2\theta + \cos^2\theta \:=\:1 \quad\Rightarrow\quad 1 - \sin^2\theta\:=\:\cos^2\theta

Substitute: x \:=\:2\sin\theta

\begin{array}{cccc}<br /> \text{Then:} &amp; \sqrt{4-x^2} \\<br /> &amp; =\;\sqrt{4-(2\sin\theta)^2} \\<br /> &amp; =\: \sqrt{4 -4\sin^2\theta} \\<br /> &amp; =\; \sqrt{4(1-\sin^2\theta)} \\<br /> &amp; =\; \sqrt{4\cos^2\theta} \\<br /> &amp; =\:2\cos\theta \end{array}

 
Do you understand the identity $\sin^2(x)+\cos^2(x)=1$ ?

Can you apply that identity to answer your question?
 
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