1. ## Does wolfram work?

Hi guys,

In the attachment you can see me trying to find the indefinite integral of a standard function:

My log book gives:

$\int \dfrac{dx}{\sqrt(a^2 - x^2)} = sin^{-1} \frac{x}{a}$

I guess I am doing something wrong in wolfram. Can anyone explain this?

2. ## Re: Does wolfram work?

Didn't you consider the possibility that they are both correct? And these are just different ways of writing the same thing?

If $\displaystyle \theta= sin^{-1}\left(\frac{x}{a}\right)$ then $\displaystyle sin(\theta)= \frac{x}{a}$. I presume you know that $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ so that $\displaystyle cos(\theta)= \sqrt{1- sin^2(\theta)}= \sqrt{1- \frac{x^2}{a^2}}= \frac{\sqrt{a^2- x^2}}{a}$.

Then $\displaystyle tan(\theta)= \frac{sin(\theta)}{cos(\theta)}= \frac{x}{\sqrt{a^2- x^2}}$.

3. ## Re: Does wolfram work?

You will also find with integration that answers can look very different, but in reality differ by only a constant. I do not have an example off the top of my head, but I remember running into examples twenty years ago when I studied integration theory.

4. ## Re: Does wolfram work?

The simplest example that I know of is
\begin{align*}
\int 2+2x \, \mathrm dx &= \int 2u \, \mathrm du &\text{where $u=1+x$} \\
2x + x^2 + c_x &= u^2 + c_u \\
&= (1+x)^2 + c_u
\end{align*}

5. ## Re: Does wolfram work?

Here's a harder one to see:

$\displaystyle \int \dfrac{1}{\sqrt{1+x^2}}dx=\text{sinh}^{-1} x+c_1 = \ln\left| x+\sqrt{1+x^2}\right|+c_2$

These actually have the same constant of integration. Nm.

6. ## Re: Does wolfram work?

The "standard" trigonometric example is $$\int 4\sin{(x)}\cos{(x)} \, \mathrm dx$$
which we can solve by trigonometric manipulation:
\begin{align*}
\int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int 2\sin{(2x)} \, \mathrm dx \\
&= -\cos{(2x)} + c_1
\end{align*}
Or by the substitution $u=\sin{(x)}$:
\begin{align*}
\int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int 4u \, \mathrm du \\
&= 2u^2 + c_2 \\
&= 2\sin^2{(x)} + c_2
\end{align*}
Or by the substitution $u=\cos{(x)}$:
\begin{align*}
\int 4\sin{(x)}\cos{(x)} \, \mathrm dx &= \int -4u \, \mathrm du \\
&= -2u^2 + c_3 \\
&= -2\cos^2{(x)} + c_3
\end{align*}

7. ## Re: Does wolfram work?

Thanks for all the examples guys. I get the picture now!