alleryIntegral ∫ 1/sqrt(1 - x²) dx =
- ln(x)
- arctan(x)
- arcsin(x) + C
- arccos(x)
Show answer and solution
Correct answer: C. arcsin(x) + C
∫ dx/sqrt(1-x²) = arcsin(x) + C. (Or -arccos(x) + C', differs by constant.) Standard integral.
AP Calculus AB Unit 1A2: Trigonometric Functions & Inverse Trig — practice questions — page 3
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Integral ∫ 1/sqrt(1 - x²) dx =Read the answer and worked solution →Integral ∫ 1/(1 + x²) dx =Read the answer and worked solution →The range of $\cos^{-1}(x)$ isRead the answer and worked solution →$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $Read the answer and worked solution →For $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equalsRead the answer and worked solution →$\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equalsRead the answer and worked solution →$\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) = $Read the answer and worked solution →The domain of the function $f(x) = \sin^{-1}(x)$ isRead the answer and worked solution →$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equalsRead the answer and worked solution →The principal value of $\tan^{-1}(1)$ isRead the answer and worked solution →For every $x \in [-1, 1]$, the value of $\sin^{-1}(x) + \cos^{-1}(x)$ isRead the answer and worked solution →Number of equivalence relations on set {a, b, c}:Read the answer and worked solution →$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equalsRead the answer and worked solution →$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equalsRead the answer and worked solution →All real $x$ satisfying $\sin^{-1}(x) + \sin^{-1}(1-x) = \cos^{-1}(x)$ areRead the answer and worked solution →If $n(A)=2$ and $n(B)=3$, the number of functions from $A$ to $B$ is:Read the answer and worked solution →If $f(x)=x^2+1$, then $f(2)$ equals:Read the answer and worked solution →If $n(A)=2$ and $n(B)=3$, the number of relations from $A$ to $B$ is:Read the answer and worked solution →In a right triangle, for an acute angle θ, cos θ is defined as:Read the answer and worked solution →The value of tan 45° is:Read the answer and worked solution →In terms of sine and cosine, tan θ equals:Read the answer and worked solution →The value of cos 0° is:Read the answer and worked solution →The fundamental trigonometric identity is:Read the answer and worked solution →The value of cos 60° is:Read the answer and worked solution →The value of sin 90° is:Read the answer and worked solution →The value of sin 30° is:Read the answer and worked solution →The reciprocal of sin θ is:Read the answer and worked solution →The domain of the function $f(x)=\dfrac{1}{x-2}$ is:Read the answer and worked solution →The range of the function $f(x)=x^2,\; x\in\mathbb{R}$ is:Read the answer and worked solution →The domain of $f(x)=\sqrt{x-1}$ is:Read the answer and worked solution →
Worked practice questions
Integral ∫ 1/(1 + x²) dx =
- 1/x
- arcsin(x)
- arctan(x) + C
- ln(1+x²)
Show answer and solution
Correct answer: C. arctan(x) + C
∫ dx/(1+x²) = arctan(x) + C. Standard integral.
Application in inverse trig substitution.
The range of $\cos^{-1}(x)$ is
- $[0, \pi]$
- $\mathbb{R}$
- $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
- $(-\pi, \pi)$
Show answer and solution
Correct answer: A. $[0, \pi]$
By the standard principal-value convention, $\cos^{-1}$ returns angles in $[0, \pi]$ — the interval where cosine is monotonically decreasing and hits every value in $[-1,1]$ exactly once.
**Option A is the range of $\sin^{-1}$ and $\tan^{-1}$, not $\cos^{-1}$.**
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Unit 3B: Implicit Differentiation and Inverse FunctionsUnit 3A: Chain Rule and Composite FunctionsUnit 1A3: Function Composition and InversesUnit 1A1: Relations, Functions and GraphsUnit 2: Differentiation and Fundamental Derivative RulesUnit 6A: Indefinite Integrals & Fundamental TheoremUnit 1B3: Asymptotic Behavior and Infinite LimitsUnit 1B1: Limit Concepts and Graphical Limits