Question

Trigonometric Substitution Make the indicated trigonometric substitution in the given algebraic expression and simplify. Assume that $\mathbf{0}\mathbf{\text{<}}\mathit{\theta }\mathbf{\text{<}}\frac{\mathbf{\pi }}{\mathbf{2}}$$\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\sqrt{\mathbf{4}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{,}\mathit{x}\mathbf{=}\mathbf{2}\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$

Short answer

The simplified expression is$\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\sqrt{\mathbf{4}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\left(co{t}^2\theta cos\theta \right)$

Step-by-step solution

Step 1. Given information

An expression $\frac{1}{x^2\sqrt{4+x^2}}$ and value for substitution as$x=2\tan \theta$

Step 2. Concept used

Substitute the value of $x$ from the question. After that, use trigonometric identities to simplify it. The solution is the result in its simplest form.

Step 3. Calculation

Now, substitute the value in $\frac{1}{x^2\sqrt{4+x^2}}$and simplify:

$\begin{array}{l}\frac{1}{x^2\sqrt{4+x^2}}=\frac{1}{4{\tan }^2\theta \sqrt{4+4{\tan }^2\theta }}\\ =\frac{1}{4{\tan }^2\theta \sqrt{4\left(1+{\tan }^2\theta \right)}}\\ \because 1+{\tan }^2\theta ={\sec }^2\theta \\ =\frac{1}{4{\tan }^2\theta \sqrt{4{\sec }^2\theta }}\\ =\frac{1}{4{\tan }^2\theta \times 2\sec \theta }\\ =\frac{1}{8{\tan }^2\theta \sec \theta }\end{array}$

Now use reciprocal identities of trigonometry,

$\begin{array}{l}\tan \theta =\frac{\sin \theta }{\cos \theta }\&\\ \sec \theta =\frac{1}{\cos \theta }\\ =\frac{1}{8\frac{{\sin }^2\theta }{{\cos }^2\theta }\times \frac{1}{\cos \theta }}\\ =\frac{1}{\frac{8{\sin }^2\theta }{{\cos }^3\theta }}\\ =\frac{{\cos }^3\theta }{8{\sin }^2\theta }\\ =\frac{{\cos }^2\theta \times \cos \theta }{8{\sin }^2\theta }\end{array}$

We can write$\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$

$=\frac{1}{8}\left({\cot }^2\theta \cos \theta \right)$

After simplification, we get-

$=\frac{1}{8}\left({\cot }^2\theta \cos \theta \right)$