Q.Evaluate: $\int \dfrac{dt}{\sqrt{3-2t-t^2}}$

The key idea is Integration by Completing the Square — rewriting the quadratic inside the square root to match the form $\frac{du}{\sqrt{a^2 - u^2}}$, which integrates to $\arcsin\left(\frac{u}{a}\right) + C$.

First, complete the square for the denominator's radicand:

The integral becomes:

Let $u = t+1$, so $du = dt$. This matches the standard form $\int \frac{du}{\sqrt{a^2 - u^2}}$ with $a = 2$: …

The key insight here is that the expression under the square root, $3 - 2t - t^2$, is a quadratic that doesn't factor nicely into a perfect square. But we can force it into a perfect square minus a constant — that's the "completing the square" technique. Once we do that, the integral becomes a standard arcsine form.

Why does this work? The derivative of $\sin^{-1}(x)$ is $\frac{1}{\sqrt{1-x^2}}$. More generally, $\frac{d}{du} \sin^{-1}\left(\frac{u}{a}\right) = \frac{1}{\sqrt{a^2 - u^2}}$. So if we can rewrite the denominator as $\sqrt{a^2 - (\text{something})^2}$, the answer is immediate.

Let's walk through it step by step.

1. Complete the square on the quadratic. Start with $3 - 2t - t^2$. Factor out the negative sign from the $t^2$ and $t$ terms:

Inside the parentheses, complete the square: $t^2 + 2t = (t+1)^2 - 1$. So:

Thus the integral becomes:

2. Recognize the standard form. The denominator is now $\sqrt{4 - (t+1)^2}$. This matches $\sqrt{a^2 - u^2}$ with $a = 2$ and $u = t+1$. The formula is:

3. Apply the formula. Here $du = dt$ (since $u = t+1$, $du = dt$), so no extra factor appears. Substituting: …

Method: Completing the square under a root, then arcsine

Use this for $\int\dfrac{dx}{\sqrt{\text{quadratic}}}$ where the quadratic has a $-x^2$ leading behaviour. Complete the square to reach $\sqrt{a^2-u^2}$ and apply arcsine.

Steps

Step 1: Arrange the quadratic and complete the square.

For $3-2t-t^2$, factor out $-1$ from the $t$-terms: $3-(t^2+2t)=3-((t+1)^2-1)=4-(t+1)^2$.

Step 2: Set $u=t+1$.

The root becomes $\sqrt{a^2-u^2}$ with $a^2=4$, i.e. $a=2$, and $du=dt$.

Step 3: Apply the arcsine standard form. …

Common Mistakes

Mistake 1: Completing the square with a sign error.

Why it's wrong: $3-2t-t^2=-(t^2+2t-3)=4-(t+1)^2$; mishandling the leading $-1$ yields $(t+1)^2-4$ and a wrong (or imaginary) form. Correct approach: factor $-1$ from the $t^2,t$ terms carefully.

Mistake 2: Forgetting to shift the variable.

Why it's wrong: the answer's argument is $\tfrac{t+1}{2}$, not $\tfrac{t}{2}$ — the completed square introduces $u=t+1$. Correct approach: substitute $u=t+1$ and keep it in the final arcsine. …