Question:
If $\int\limits^x_1 \frac{dt}{|t\, \sqrt{t^2 - 1}} = \frac{\pi}{6}$ , then $x$ can be equal to
If $\int\limits^x_1 \frac{dt}{|t\, \sqrt{t^2 - 1}} = \frac{\pi}{6}$ , then $x$ can be equal to
Updated On: Jun 18, 2022
- $\frac{ 2}{\sqrt{3}}$
- $\sqrt{3}$
- $2$
- None of these
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The Correct Option is A
Solution and Explanation
$\int\limits_{1}^{x} \frac{d t}{|t| \sqrt{t^{2}-1}}=\frac{\pi}{6}$
$\Rightarrow \left[\sec ^{-1} t\right]_{1}^{x}=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x-\sec ^{-1} 1=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x- 0=\frac{\pi}{6}$
$\Rightarrow x=\sec \frac{\pi}{6}$
$\Rightarrow x=\frac{2}{\sqrt{3}}$
$\Rightarrow \left[\sec ^{-1} t\right]_{1}^{x}=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x-\sec ^{-1} 1=\frac{\pi}{6}$
$\Rightarrow \sec ^{-1} x- 0=\frac{\pi}{6}$
$\Rightarrow x=\sec \frac{\pi}{6}$
$\Rightarrow x=\frac{2}{\sqrt{3}}$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
