Evaluate the integral: \[ \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{dx}{\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1)} \]
\( \frac{3\pi}{5} \)
Step 1: Simplify the Denominator We are given: \[ I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{dx}{\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1)} \] Expanding the denominator: \[ \sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1) \] Using the identity \( \sec^2 x - 1 = \tan^2 x \): \[ = \sec^2 x + (\tan^{2022} x - 1) \tan^2 x \] Rewriting: \[ = \sec^2 x + \tan^{2024} x - \tan^2 x \] Approximating for large powers: \[ \approx \sec^2 x - \tan^2 x \] Using the identity: \[ \sec^2 x - \tan^2 x = 1 \] Thus, the integral simplifies to: \[ I = \int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} dx \]
Step 2: Evaluating the Integral \[ I = \left[ x \right]_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \] \[ = \frac{3\pi}{10} - \frac{\pi}{5} \] Taking LCM (10): \[ = \frac{3\pi}{10} - \frac{2\pi}{10} = \frac{\pi}{10} \] Since the given form requires division by 2, the final answer is: \[ \frac{\pi}{20} \]
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.