Compute the integral $\int\limits_0^1 \frac{3x}{\sqrt{4-3x^2}} dx $?

Question

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}

Answer 1

Hint:

If you let $u=4-3x^2,$ then all will be fine and dandy.

Answer 2

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