1. ## integral

Hi,

I'm trying to integrate the following, I can see that if the exponent of the function was negative 1/2 it would integrate to cosh. I'm not sure how to use that information:

$\displaystyle \int \sqrt{(4x^2-1)} \, \mathrm{d}x.$

thanks

2. ## Re: integral

Let $x = \dfrac{1}{2}\sec \theta$. Then $dx = \dfrac{1}{2}\sec \theta \tan \theta d\theta$

Plugging in:

\begin{align*}\displaystyle \int \sqrt{4x^2-1} dx & = \dfrac{1}{2}\int \sqrt{\sec^2 \theta - 1} \sec \theta \tan \theta d\theta \\ & = \dfrac{1}{2}\int \sqrt{\tan^2 \theta}\sec \theta \tan \theta d\theta \\ & = \dfrac{1}{2}\int \sin \theta \tan \theta \sec^2 \theta d\theta\end{align*}

Let $u = \sin \theta, dv = \tan \theta \sec^2 \theta d\theta$. Then $w = \tan \theta, dw = \sec^2 \theta d\theta$, so $dv = wdw$. Integrating, we get $v = \dfrac{1}{2}w^2 = \dfrac{1}{2}\tan^2 \theta$ and $du = \cos \theta d\theta$.

So, plugging in, you get:

\begin{align*}\displaystyle \dfrac{1}{2}\int \sin \theta \tan \theta \sec^2 \theta d\theta & = \dfrac{1}{4}\sin \theta \tan^2 \theta - \dfrac{1}{4}\int \cos \theta \tan^2 \theta d\theta \\ & = \dfrac{1}{4}\sin \theta \tan^2 \theta - \dfrac{1}{4} \int \dfrac{\sin^2 \theta}{\cos \theta} d\theta \\ & = \dfrac{1}{4}\sin \theta \tan^2 \theta - \dfrac{1}{4} \int \dfrac{1-\cos^2 \theta}{\cos \theta} d\theta \\ & = \dfrac{1}{4}\sin \theta \tan^2 \theta - \dfrac{1}{4} \int \sec \theta d\theta + \dfrac{1}{4}\int \cos \theta d\theta \\ & = \dfrac{1}{4}\sin \theta \tan^2 \theta - \dfrac{1}{4} \ln \left| \sec \theta + \tan \theta \right| + \dfrac{1}{4}\sin \theta + C \\ & = \dfrac{1}{4} \left( \sin \theta(\tan^2 \theta +1) - \ln \left|\sec \theta + \tan \theta\right|\right) + C \\ & = \dfrac{1}{4}\left(\sin \theta \sec^2 \theta - \ln \left|\sec \theta + \tan \theta \right| \right) + C\end{align*}

Next, you need to substitute back in. Since $x = \dfrac{1}{2}\sec \theta$, you have $\sec \theta = 2x$. Since $\sec \theta = \dfrac{\text{hyp}}{\text{adj}}$, you can figure out the other trig functions: $\text{opp} = \sqrt{4x^2-1}, \text{adj} = 1, \text{hyp} = 2x$. Then, $\sin \theta = \dfrac{\sqrt{4x^2-1}}{2x}, \tan \theta = \sqrt{4x^2-1}$.

So, this gives:

\begin{align*}\dfrac{1}{4}\left(\sin \theta \sec^2 \theta - \ln \left|\sec \theta + \tan \theta \right| \right) + C & = \dfrac{1}{4}\left(\dfrac{\sqrt{4x^2-1}}{2x}(2x)^2 - \ln \left|2x + \sqrt{4x^2-1}\right|\right) + C \\ & = \dfrac{1}{4}\left(2x\sqrt{4x^2-1} - \ln\left|2x + \sqrt{4x^2-1} \right| \right) + C\end{align*}

3. ## Re: integral

you're not gettin paid enough for this....

4. ## Re: integral

Alternatively, you could do it with a hyperbolic sub.

$$x = \tfrac{1}{2}\cosh u:$$
\begin{align*} \int \sqrt{4x^2 - 1} \ \text{d}x & = \frac{1}{2} \int \sqrt{ \cosh^2 u - 1} \sinh u \ \text{d}u \\ & = \frac{1}{2} \int \sinh^2 u \ \text{d}u \\ & = \frac{1}{4} \int \cosh 2u - 1 \ \text{d}u \\ & = \frac{1}{4} \left( \tfrac{1}{2} \sinh 2u - u \right) + C \\ & = \frac{1}{4}\big( \sinh u \cosh u - u\big) + C \end{align*}

Now use $x = \tfrac{1}{2}\cosh u \iff \cosh u = 2x, \ \sinh u = \sqrt{4x^2 - 1}$ to get:

$$\int \sqrt{4x^2 - 1}\ \text{d}x = \frac{1}{4}\Big(2x \sqrt{4x^2 - 1} - \text{arcosh} 2x\Big) + C = \frac{1}{4} \Big(2x \sqrt{4x^2 - 1} - \ln \left| 2x + \sqrt{4x^2 - 1} \right|\Big) + C$$