Substitute $x=a\sin{\theta}$ into $\frac{ax^2}{\sqrt{a^2-x^2}}$ and simplify

Question

Question

Substitute $x=a\sin{\theta}$ into $\frac{ax^2}{\sqrt{a^2-x^2}}$

My thoughts

So when i substituted, i got the following:

$\frac{a*a^2\sin^2{\theta}}{\sqrt{a^2-a^2\sin^2{\theta}}}$

But then I didnt know how to simplify from there

Answer 1

$$\frac{a^3\sin^2\theta}{\sqrt {a^2-a^2\sin^2\theta}}= \pm\frac{a^2\sin^2\theta}{\sqrt{1-\sin^2\theta}}=\pm\frac{a^2\sin^2\theta}{\cos\theta}=\pm a^2\sin\theta\tan\theta$$ I think this is what is needed (I used the trig identity $\sin^2\theta+\cos^2\theta=1$)

Answer 2

$x=a\sin{\theta}$ into $\frac{ax^2}{\sqrt{a^2-x^2}}$

$$\frac{ax^2}{\sqrt{a^2-x^2}}$$

$$\frac{a^3 \sin^2(\theta)}{\sqrt{a^2-a^2\sin^2(\theta)}}$$

$$\frac{a^3 \sin^2(\theta)}{\sqrt{a^2(1- \sin^2(\theta))}}$$

$$\frac{a^3 \sin^2(\theta)}{\sqrt{a^2\cos^2(\theta)}}$$

$$ \frac{a^3 \sin^2(\theta)}{a\cos(\theta)}$$

$$ \frac{a^2 \sin^2(\theta)}{\cos(\theta)}$$

Assuming $a>0$ and $\cos(\theta)>0$

Answer 3

$$\text{Z}=\frac{\text{a}\cdot\left(\text{a}\cdot\sin\theta\right)^2}{\sqrt{\text{a}^2-\left(\text{a}\cdot\sin\theta\right)^2}}=\frac{\text{a}\cdot\text{a}^2\cdot\sin^2\theta}{\sqrt{\text{a}^2-\text{a}^2\cdot\sin^2\theta}}=\frac{\text{a}^3\cdot\sin^2\theta}{\sqrt{\text{a}^2\left(1-\sin^2\theta\right)}}=$$ $$\frac{\text{a}^3\cdot\sin^2\theta}{\left|\text{a}\right|\cdot\sqrt{1-\sin^2\theta}}=\frac{\text{a}^3}{\left|\text{a}\right|}\cdot\frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}}\tag1$$

Now, use:

• $$\sin^2\theta=\frac{1-\cos2\theta}{2}\tag2$$
• $$\cos^2\theta=1-\frac{1-\cos2\theta}{2}\tag3$$

So, we get:

$$\text{Z}=\frac{\text{a}^3}{\left|\text{a}\right|}\cdot\frac{\frac{1-\cos2\theta}{2}}{\sqrt{1-\frac{1-\cos2\theta}{2}}}=\frac{1}{2}\cdot\frac{\text{a}^3}{\left|\text{a}\right|}\cdot\frac{1-\cos2\theta}{\sqrt{\cos^2\theta}}=\frac{1}{2}\cdot\frac{\text{a}^3}{\left|\text{a}\right|}\cdot\frac{1-\cos2\theta}{\left|\cos\theta\right|}\tag4$$