Chapter 5: Q. 83 (page 465)

Solve given definite integral.
$\int_{1 / 4}^{1 / 2} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$

Short Answer

Expert verified

$- \sqrt{3} + \sqrt{15}$

Step by step solution

Step1. Given Information

The integral is as follows.

$\int_{\frac{1}{4}}^{\frac{1}{2}} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$

The objective is to solve the integral.

Step2. Assumption

To solve the integral, let $x = \sin (u)$

Now, derivation of above equation is solved below.

$x = \sin (u)$

$d x = \cos (u) d u$

Step3. Solution

Now use the identity, $- \sin^{2} (u) + 1 = \cos^{2} (u)$

So,

$\frac{1}{x^{2} \sqrt{1 - x^{2}}} = \frac{1}{\sqrt{- \sin^{2} (u) + 1} \sin^{2} (u)}$

$\begin{aligned} = \frac{1}{\sqrt{\cos^{2} (u)} \sin^{2} (u)} \\ = \frac{1}{\cos (u) \sin^{2} (u)} \end{aligned}$

Step4. Solution

The integral is solved below

$\begin{aligned} \int \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x \\ = \int (\frac{1}{\sin^{2} (u)}) d u \\ = \int \csc^{2} (u) d u \\ = - \cot (u) \\ = - \cot (\sin^{- 1} (x)) \\ = - \frac{1}{x} \sqrt{- x^{2} + 1} \end{aligned}$

Step5. Solution

The definite integral is solved below.

$\int_{\frac{1}{4}}^{\frac{1}{2}} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$

$= {[- \frac{1}{x} \sqrt{- x^{2} + 1}]}_{\frac{1}{4}}^{\frac{1}{2}}$

$= [- \frac{1}{(\frac{1}{2})} \sqrt{- {(\frac{1}{2})}^{2} + 1}] - [- \frac{1}{(\frac{1}{4})} \sqrt{- {(\frac{1}{4})}^{2} + 1}]$

$= - \sqrt{3} + \sqrt{15}$

Therefore, the value is $= - \sqrt{3} + \sqrt{15}$

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Solve given definite integral.
$\int_{1 / 4}^{1 / 2} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$

Short Answer

Step by step solution

Step1. Given Information

Step2. Assumption

Step3. Solution

Step4. Solution

Step5. Solution

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Step by step solution

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Step2. Assumption

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Step5. Solution

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Solve given definite integral.
$\int_{1 / 4}^{1 / 2} \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$