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Indefinite Integrals · Question 55 of 90

Question 55

MediumOriginal question

Read the two statements and choose the correct option.

I. ∫dxx2+a2=1atan⁡−1xa+C\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C.

II. ∫dxa2−x2=1asin⁡−1xa+C\int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \frac{1}{a}\sin^{-1}\frac{x}{a} + C.

Options for question 55

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. I: ddx(1atan⁡−1xa)=1a⋅1/a1+x2/a2=1x2+a2\frac{d}{dx}\left(\frac{1}{a}\tan^{-1}\frac{x}{a}\right) = \frac{1}{a} \cdot \frac{1/a}{1 + x^{2}/a^{2}} = \frac{1}{x^{2} + a^{2}}. True.

  2. II: ddxsin⁡−1xa=1a2−x2\frac{d}{dx}\sin^{-1}\frac{x}{a} = \frac{1}{\sqrt{a^{2} - x^{2}}} already, so the extra 1a\frac{1}{a} is wrong. False.

Concept
The table of particular integrals
Typical time
90 s
Your time
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The rule this testsIntegrals of particular functions
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Ten questions from Indefinite Integrals, at your own pace.

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Ten questions from Indefinite Integrals, at your own pace.