# Use the indicated substitution to evaluate the integral. Let x= 4\tan t. \int...

## Question:

Use the indicated substitution to evaluate the integral.

Let {eq}x= 4\tan t. {/eq}

{eq}\int \frac{dx}{x^2\sqrt{x^2+16}}, {/eq} [1/2, 1]

## Definite Integral:

The definite integral of a function is closely related to the antiderivative and indefinite integral of a function. In calculus, the definite integrals can be find by first evaluating the value of the indefinite integral and then substituting the limits to find the required value.

Let {eq}I = \int \frac{dx}{x^2\sqrt{x^2+16}} {/eq}

We can evaluate the above integral by using substitution method.

Let {eq}x= 4\tan t \ \Rightarrow dx = 4\sec^2 t \ dt {/eq}

Thus,

{eq}\begin{align*} \displaystyle I & = \displaystyle \int \frac{dx}{x^2\sqrt{x^2+16}}\\ & = \displaystyle \int \frac{4\sec^2 t \ dt}{16 \tan^2 t \sqrt{16\tan^2 t+16}}\\ & = \displaystyle \int \frac{4\sec^2 t \ dt}{64 \tan^2 t \sqrt{\tan^2 t+1}}\\ & = \displaystyle \int \frac{\sec^2 t \ dt}{16 \tan^2 t \sec t}\\ & = \displaystyle \int \frac{\sec t \ dt}{16 \tan^2 t}\\ & = \displaystyle \frac{1}{16} \int \frac{\cos t \ dt}{\sin^2 t}\\ \end{align*} {/eq}

Now, let {eq}u = \sin t \Rightarrow du = \cos t \ dt {/eq}

Thus,

{eq}\begin{align*} \displaystyle I & = \displaystyle \frac{1}{16} \int \frac{\cos t \ dt}{\sin^2 t}\\ & = \displaystyle \frac{1}{16} \int \frac{du}{u^2}\\ & = \displaystyle \frac{1}{16} \times \frac{-1}{u}\\ & = \displaystyle \frac{-1}{16\sin t}\\ & = \displaystyle \frac{-1}{16\sin (\tan^{-1} (x/4)}\\ & = \displaystyle \frac{-\sqrt{x^2+16}}{16x}\\ \end{align*} {/eq}

So,

{eq}\begin{align*} \displaystyle I & = \displaystyle \int_{1/2}^{1}\frac{dx}{x^2\sqrt{x^2+16}}\\ & = \displaystyle \left(\frac{-\sqrt{x^2+16}}{16x}\right)_{1/2}^{1}\\ & = \displaystyle \left(\frac{-\sqrt{1+16}}{16} - \frac{-\sqrt{(1/4)+16}}{8} \right)\\ & = \displaystyle \left(\frac{-\sqrt{17}}{16} + \frac{\sqrt{65}}{16} \right)\\ \end{align*} {/eq}