Find the value of x for which integral from 1 to x dt/(t

Find the value of $x$ for which $\int_1^x \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{3}$.

We will first find the antiderivative, then solve for $x$.

Step 1: Find the antiderivative of $\frac{1}{t\sqrt{t^2-1}}$

This is a standard integral form. Use the substitution $t = \sec\theta$, so $dt = \sec\theta\tan\theta \, d\theta$, and:

\[\sqrt{t^2-1} = \sqrt{\sec^2\theta - 1} = \sqrt{\tan^2\theta} = \tan\theta \quad \left(\theta \in \left(0, \frac{\pi}{2}\right)\right)\]

Therefore:

\[\int \frac{dt}{t\sqrt{t^2-1}} = \int \frac{\sec\theta\tan\theta \, d\theta}{\sec\theta \cdot \tan\theta} = \int d\theta = \theta + C\]

Since $t = \sec\theta$, this is $\theta = \operatorname{arcsec}(t)$, i.e.

\[\int \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{t}\right) + C\]

Verification:

Let $F(t) = \arccos\left(\frac{1}{t}\right)$. Then $F'(t) = -\frac{1}{\sqrt{1-1/t^2}}\cdot\left(-\frac{1}{t^2}\right) = \frac{1}{t^2\sqrt{1-1/t^2}} = \frac{1}{t\sqrt{t^2-1}}$.

Step 2: Evaluate the definite integral

\[\int_1^x \frac{dt}{t\sqrt{t^2-1}} = \left[\arccos\left(\frac{1}{t}\right)\right]_1^x = \arccos\left(\frac{1}{x}\right) - \arccos(1) = \arccos\left(\frac{1}{x}\right)\]

Step 3: Set the integral equal to $\frac{\pi}{3}$ and solve

\[\arccos\left(\frac{1}{x}\right) = \frac{\pi}{3} \quad\Rightarrow\quad \frac{1}{x} = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \quad\Rightarrow\quad x = 2\]

Verification:

\[\int_1^2 \frac{dt}{t\sqrt{t^2-1}} = \arccos\left(\frac{1}{2}\right) - \arccos(1) = \frac{\pi}{3} - 0 = \frac{\pi}{3} \quad \checkmark\]

Check the domain:

For the integral to be defined we need $t>1$ (so that $t^2-1>0$) and the integral to converge; since $x=2>1$, this is valid.

Geometric interpretation:

The integral $\int_1^x \frac{dt}{t\sqrt{t^2-1}}$ represents the angle (in radians) whose secant is $x$. When this angle equals $\frac{\pi}{3}$ ($60°$), we have $\sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos(\pi/3)} = 2$, confirming our answer.

Final answer:

\[\boxed{x = 2}\]

Additional note:

The antiderivative can also be written as $\operatorname{arcsec}(t)+C$, $\arccos\left(\frac1t\right)+C$, or $\arcsin\left(\frac{\sqrt{t^2-1}}{t}\right)+C$ — all differ only by a constant and lead to the same result.