MHB Solving an integral with trigonometric substitution

[tmt1]

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?

 

[soroban]

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}

 

[Greg]

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

Can you apply that identity to answer your question?