# Solve and show all steps. Do not give the answer in terms of hyperbolic functions. Compute ...

## Question:

Solve and show all steps. Do not give the answer in terms of hyperbolic functions.

Compute {eq}\displaystyle\int_0^1 \frac{dx}{\sqrt {x^2 + 2x}}. {/eq}

## Trigonometric Substitution:

If we have a square root in the denominator of an integrand, and it does not contain the sum or difference of squares, but rather a square and a linear term, then we can complete the square inside the square root. Then we factor the perfect square trinomial, and apply the appropriate trigonometric substitution.

First, it is important to notice that this is an improper integral, since the integrand does not exist when {eq}x = 0. {/eq}

{eq}\begin{eqnarray*}\displaystyle\int_0^1 \frac{dx}{\sqrt {x^2 + 2x}} & = & \displaystyle\lim_{a \to 0^+} \int_a^1 \frac{dx}{\sqrt{x^2 + 2x}} \\ \\ & = & \displaystyle\lim_{a \to 0^+} \int_a^1 \frac{dx}{\sqrt{(x^2 + 2x + 1) - 1}} \\ \\ & = & \displaystyle\lim_{a \to 0^+} \int_a^1 \frac{dx}{\sqrt{(x + 1)^2 - 1}} \end{eqnarray*} {/eq}

Now we define {eq}x + 1 = \sec \theta, {/eq} so {eq}dx = \sec \theta \tan \theta \: d\theta. {/eq} When {eq}x = a {/eq} we have {eq}a + 1 = \sec \theta, {/eq} so {eq}\theta = \sec^{-1} (a + 1). {/eq} When {eq}x = 1 {/eq} we have {eq}\sec \theta = 2, {/eq} so {eq}\theta = \displaystyle\frac{\pi}3. {/eq} Therefore

{eq}\begin{eqnarray*}\displaystyle\lim_{a \to 0^+} \int_a^1 \frac{dx}{\sqrt{(x + 1)^2 - 1}} & = & \displaystyle\lim_{a \to 0^+} \int_{\sec^{-1} (a + 1)}^{\pi/3} \frac{\sec \theta \tan \theta}{\sqrt{\sec^2 \theta - 1}} \: d\theta \\ \\ & = & \displaystyle\lim_{a \to 0^+} \int_{\sec^{-1} (a + 1)}^{\pi/3} \frac{\sec \theta \tan \theta}{\sqrt{\tan^2 \theta}} \: d\theta \\ \\ & = & \displaystyle\lim_{a \to 0^+} \int_{\sec^{-1} (a + 1)}^{\pi/3} \sec \theta \: d\theta \\ \\ & = & \displaystyle\lim_{a \to 0^+} \ln |\sec \theta + \tan \theta| \: \biggr|_{\sec^{-1} (a + 1)}^{\pi/3} \\ \\ & = & \displaystyle\lim_{a \to 0^+} \ln \left| \sec \displaystyle\frac{\pi}3 + \tan \frac{\pi}3 \right| - \ln \left| \sec \sec^{-1} (a + 1) + \tan \sec^{-1} (a + 1) \right| \end{eqnarray*} {/eq}

Now, if {eq}\theta = \sec^{-1} (a + 1), {/eq} then {eq}\sec \theta = a + 1. {/eq} A triangle for this problem appears below:

We can see from the triangle that {eq}\tan \theta = \sqrt{a^2 + 2a}. {/eq} Therefore

{eq}\begin{eqnarray*}\displaystyle\lim_{a \to 0^+} \ln \left| \sec \displaystyle\frac{\pi}3 + \tan \frac{\pi}3 \right| - \ln \left| \sec \sec^{-1} (a + 1) + \tan \sec^{-1} (a + 1) \right| & = & \displaystyle\lim_{a \to 0^+} \ln |2 + \sqrt 3| - \ln \left| a + 1 - \sqrt{a^2 + 2a} \right| \\ \\ & = & \ln (2 + \sqrt 3) \end{eqnarray*} {/eq}