Compute the integral $\int\limits_0^1 \frac{3x}{\sqrt{4-3x^2}} dx $? I am struggling to compute the following equation.
\begin{equation}
    \displaystyle\int_0^1 \dfrac{3x}{\sqrt{4-3x^2}} dx 
\end{equation}
We are expected to use u-substitution, but I'm stuck and don't know how to proceed.
I don't know how to compute the integral of an equation in the form of:
\begin{equation}
    \frac{1}{\sqrt{1-ax^{2}}}
\end{equation}
 A: Hint:
If you let $u=4-3x^2,$ then all will be fine and dandy.
A: In general, we would like to simplify the equation by use of u-substitution.
That being said, it would be smart to choose a value for $u$ that simplifies the equation to one that we know how to solve.
There are many options for $u$ in this case.
@Allawonder pointed out that $u=4-3x^{2}$ might be a good place to start, so let's use that.
\begin{align}
    \int_0^1 \frac{3x}{\sqrt{4-3x^2}} dx \\
    3\int_0^1 \frac{x}{\sqrt{4-3x^2}} dx \\
    u = 4-3x^2 \tag{declare} \\
    du = -6x dx \\
    -\frac{1}{6}du = x dx
\end{align}
Now, let's find the new values for $a$ and $b$ for this u-substitution.
\begin{align}
    a = 4 \\
    b = 1 \\
\end{align}
Now substitute the computed values:
\begin{align}
    &=3\int_4^1 -\frac{1}{6\sqrt{u}} du \\
    &=-\frac{1}{2}\int_4^1 \frac{1}{\sqrt{u}} du \\
    &=\frac{1}{2}\int_1^4 \frac{1}{\sqrt{u}} du \\
\end{align}
Now we have something that looks more familiar.
\begin{align}
    &=\frac{1}{2}\left[ 2\sqrt{u}\right]^{1}_{4} \\
    &=\left[ \sqrt{u} \right]^{1}_{4} \\
    &=\left[ \sqrt{(4)} - \sqrt{(1)}\right]
    &=1
\end{align}
And there we go!
Hope this helped
