Q. 50i Solve the integral鈭玿2-42xdx... [FREE SOLUTION]

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Chapter 5: Q. 50i (page 465)

Solve the integral
$鈭� \frac{\sqrt{x^{2} - 4}}{2 x} d x$

Short Answer

Expert verified

$\frac{1}{2} \sqrt{x - 2} \sqrt{x + 2} - \tan^{- 1} (\frac{1}{2} \sqrt{x - 2} \sqrt{x + 2}) + C$

Step by step solution

Step 1. Given

$鈭� \frac{\sqrt{x^{2} - 4}}{2 x} d x$

Step 2. Simplifying the equation

$To solve the integral, let x = 2 \cosh (u) Now, derivation of the above equation is solved below . x = 2 \cos h (u) d x = 2 \sin h (u) d u Now, use the identity c o s h^{2} (u) - 1 = s i n h^{2} (u) S o, \frac{\sqrt{(x^{2} - 4)}}{2 x} = \frac{1}{x} \sqrt{(x^{2} - 4)} = \frac{\sqrt{4 \cos h^{2} (u) - 4}}{2 c o s h (u)} = \frac{\sqrt{\cos h^{2} (u) - 1}}{c o s h (u)} = \frac{\sqrt{\sin h^{2} (u)}}{\cos h (u)}$

Step 3. Solving the integral

$鈭� \frac{\sqrt{x^{2} - 4}}{2 x} d x = \frac{1}{2} 鈭� \frac{2 \sin h^{2} (u)}{c o s h (u)} d u = 鈭� \frac{2 \sin h^{2} (u)}{c o s h (u)} d u = 鈭� \frac{\sin h^{2} (u) \cos h (u)}{s i n h^{2} (u) + 1} d u = 鈭� \frac{v^{2}}{v^{2} + 1} d v [v = s i n h (u), d v = c o s h (u)] 鈭� 1 d v - 鈭� \frac{1}{v^{2} + 1} d v = v - t a n^{- 1} (v) + C = s i n h (u) - t a n^{-} (s i n h (u)) + C = s i n h (c o s h^{-} 1 (\frac{x}{2}) - t a n^{-} 1 (s i n h (c o s h^{-} 1 (\frac{x}{2}))) + C = \frac{1}{2} \sqrt{x - 2} \sqrt{x + 2} - \tan^{- 1} (\frac{1}{2} \sqrt{x - 2} \sqrt{x + 2}) + C$

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91影视

Solve the integral
$鈭� \frac{\sqrt{x^{2} - 4}}{2 x} d x$

Short Answer

Step by step solution

Step 1. Given

Step 2. Simplifying the equation

Step 3. Solving the integral

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91影视

Solve the integral鈭�x2-42xdx

Short Answer

Step by step solution

Step 1. Given

Step 2. Simplifying the equation

Step 3. Solving the integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Discrete Mathematics

Calculus

Geometry

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Solve the integral
$鈭� \frac{\sqrt{x^{2} - 4}}{2 x} d x$