Chapter 5: Q. 83 (page 465)

Solve given definite integral.
${鈭�}_{1 / 4}^{1 / 2} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$

Short Answer

Expert verified

$鈭� \sqrt{3} + \sqrt{15}$

Step by step solution

Step1. Given Information

The integral is as follows.

${鈭�}_{\frac{1}{4}}^{\frac{1}{2}} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$

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,

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

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

Step4. Solution

The integral is solved below

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

Step5. Solution

The definite integral is solved below.

${鈭�}_{\frac{1}{4}}^{\frac{1}{2}} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$

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

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

$= 鈭� \sqrt{3} + \sqrt{15}$

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

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Solve given definite integral.
${鈭�}_{1 / 4}^{1 / 2} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$

Short Answer

Step by step solution

Step1. Given Information

Step2. Assumption

Step3. Solution

Step4. Solution

Step5. Solution

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91影视

Solve given definite integral.鈭�1/41/2鈥�1x21鈭�x2dx

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Step by step solution

Step1. Given Information

Step2. Assumption

Step3. Solution

Step4. Solution

Step5. Solution

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Most popular questions from this chapter

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Solve given definite integral.
${鈭�}_{1 / 4}^{1 / 2} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$