# Manipulating equations.

#### shamieh

##### Active member
I know this is easy but im getting hung up on this problem for some reason.

Suppose we have $$\displaystyle \sqrt{8}sin\theta$$ and we plug in $$\displaystyle 2$$ for $$\displaystyle \theta$$.

$$\displaystyle \sqrt{8}sin(2).$$

how are they then coming to the conclusion of $$\displaystyle \sin\theta = \frac{1}{\sqrt{2}}$$

That makes NO sense! We just plugged a $$\displaystyle 2$$ in for $$\displaystyle \theta$$! Where does the extra $$\displaystyle \theta$$ come from?!

#### Ackbach

##### Indicium Physicus
Staff member
I'm afraid you're going to have to give us some context for this question. Can you please give us the original problem statement, word-for-word?

(As a side note, you can't go from an expression to an equation, so yes, I'm very puzzled by that transition myself.)

#### shamieh

##### Active member
My teacher has left out some weird step that I am missing! Originally I am solving this $$\displaystyle 2$$ $$\displaystyle \int^2_0 \sqrt{8 - x^2} - \frac{1}{2}x^2 dx$$

Ignore the second, that is easy so I split it up into two different integrals. Here is what I have so far, notice I am only doing the first one.

$$\displaystyle 2 \int^2_0 \sqrt{8 -x^2} dx$$

$$\displaystyle 2 \int^2_0 \sqrt{8 - \sin^2\theta} * \sqrt{8}\cos\theta d\theta$$

Also note that I have: $$\displaystyle 1 - \sin^2\theta = cos^2\theta$$
$$\displaystyle 8 - 8\sin^2\theta = 8\cos^2\theta$$
$$\displaystyle x^2 = 8sin^2\theta$$
$$\displaystyle x = \sqrt{8}\sin\theta$$
$$\displaystyle dx = \sqrt{8}\cos\theta \, d\theta$$

So now since I've done the substitution i need to update the limits... So my teacher said ok plug in the old limits into x.. Which I understand the first limit. He plugs in a 0. $$\displaystyle \sqrt{8}sin(0)$$= 0 thus the lower limit is 0 but for the upper limit he plugs in a 2 ....$$\displaystyle \sqrt{8}sin(2)$$ and somehow gets $$\displaystyle 2 = \sqrt{8}sin\theta$$ which becomes
$$\displaystyle sin\theta = \frac{1}{\sqrt{2}}$$

Like what am I missing here??

#### Ackbach

##### Indicium Physicus
Staff member
Ah, thanks much for that background. Let's consider this integral:
$$\int_{0}^{2} \sqrt{8-x^{2}} \, dx.$$
I would indeed do a trig substitution. I think you have a bit of ambiguity in your substitution. Draw a right triangle, angle $\theta$, hypotenuse $\sqrt{8}$, and opposite side $x$. Then the adjacent side is $\sqrt{8-x^{2}}$. You have that
\begin{align*}
\frac{x}{ \sqrt{8}}&= \sin(\theta) \\
\frac{dx}{ \sqrt{8}}&= \cos(\theta) \, d\theta \\
\frac{ \sqrt{8-x^{2}}}{ \sqrt{8}}&= \cos( \theta)
\end{align*}
That will enable you do substitute in for the integrand and the differential. Now, the limits: $0$ and $2$ are $x$ values, because that is the variable in the differential. To get to $\theta$ values, you have to solve $0= \sqrt{8} \sin(\theta)$ for $\theta$, as well as $2= \sqrt{8} \sin( \theta)$ for $\theta$. There's no plugging $0$ or $2$ into the trig functions, because that's a category mistake: $0$ and $2$ are $x$-values, not $\theta$-values. Does that answer your question?

Thanks so much!

#### shamieh

##### Active member
So is my work correct?

$$\displaystyle 2 \int ^\frac{\pi}{4}_0$$

$$\displaystyle 2 \int ^\frac{\pi}{4}_0 \sqrt{8} cos\theta * \sqrt{8}cos\theta$$

$$\displaystyle 2 \int ^\frac{\pi}{4}_0 8 cos^2\theta$$

then I brought the 8 out and got

$$\displaystyle 16 [ \theta - \frac{1}{2} sin2\theta] |^{\pi/4}_0$$

#### shamieh

##### Active member
I don;'t understand what ive done wrong. How are they getting $$\displaystyle 2\pi + \frac{4}{3}$$ as the final answer?

#### Ackbach

##### Indicium Physicus
Staff member
Yes, I think your re-worked integral is correct. WolframAlpha confirms your result, both the original and the re-worked integral.