# 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.

## Answer and Explanation:

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}

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