Chapter 5: Q4P (page 242)

$\int \sqrt{1 + " " a "^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + " " a "^{2} x^{2}} + \frac{1}{2 a} s i n h^{- 1} \sqrt{a^{2} + b^{2}}$

Short Answer

Expert verified

The required solution is $\int \sqrt{1 + x^{2} a^{2})} d x = \frac{1}{2 a} [\sin h^{- 1} (a x) + 1 + x^{2} a^{2} x a] + c$

Step by step solution

Definition of definite integral

A function that practices the antiderivative of another function is called an indefinite integral. It can be represented graphically as an integral symbol, a function, and finally a dx. The indefinite integral is a more straightforward way of expressing the antiderivative.

Using the substitution

Make use of the substitute $x = (\frac{1}{2}) \sin h u$

role="math" localid="1658836295640" $\int \sqrt{1 + x^{2} a^{2}} d x = \int \sqrt{1 + \sin h^{2} u} \frac{1}{a} \cos h u d u = \frac{1}{a} \int \cos h^{2} u d u = \frac{1}{a} \int \frac{1}{2} (\cos h (2 u) + 1) d u = \frac{1}{2 a} [u + \int \cos h (2 u) d u] = \frac{1}{2 a} [u + \frac{1}{2} \sin h (2 u)] + C = \frac{1}{2 a} [u + \cos h u \sin h u] + C = \frac{1}{2} [\sin h^{- 1} (a x) + \sqrt{1 + \sin h^{2} u x a}] + C = \frac{1}{2} [\sin h^{- 1} (a x) + \sqrt{1 + x^{2} a^{2} x a}] + C$

Now, Apply role="math" localid="1658836375727" $c o s h^{2} " x - s i n h^{2} x = 1 a n d c o s h^{2} x = (1 + c o s h (2 x)) / 2 .$

Therefore, the answer to the question is

$\int \sqrt{1 + x^{2} a^{2}} d x = \frac{1}{2 a} [\sin h^{- 1} (a x) + \sqrt{1 + x^{2} a^{2}} x a] + C$

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$\int \sqrt{1 + " " a "^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + " " a "^{2} x^{2}} + \frac{1}{2 a} s i n h^{- 1} \sqrt{a^{2} + b^{2}}$

Short Answer

Step by step solution

Definition of definite integral

Using the substitution

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∫1+""a"2x2dx=x21+""a"2x2+12asinh-1a2+b2

Short Answer

Step by step solution

Definition of definite integral

Using the substitution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

Work Energy and Power

Classical Mechanics

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Engineering Physics

Study anywhere. Anytime. Across all devices.

$\int \sqrt{1 + " " a "^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + " " a "^{2} x^{2}} + \frac{1}{2 a} s i n h^{- 1} \sqrt{a^{2} + b^{2}}$