int_0^a sqrt(a^2-x^2)dx=?

প্রশ্ন: \( \int_0^a \sqrt{a^2 - x^2} \, dx = ? \) 🧐

সমাধান:

1. ধরি, \( x = a \sin(\theta) \) সুতরাং, \( dx = a \cos(\theta) \, d\theta \) 🤩
2. যখন \( x = 0 \), \( \theta = 0 \) এবং যখন \( x = a \), \( \theta = \frac{\pi}{2} \) 🥳
3. অতএব, \[ \int_0^a \sqrt{a^2 - x^2} \, dx = \int_0^{\frac{\pi}{2}} \sqrt{a^2 - a^2 \sin^2(\theta)} \cdot a \cos(\theta) \, d\theta \]
4. \[ = \int_0^{\frac{\pi}{2}} \sqrt{a^2(1 - \sin^2(\theta))} \cdot a \cos(\theta) \, d\theta \]
5. \[ = \int_0^{\frac{\pi}{2}} a \cos(\theta) \cdot a \cos(\theta) \, d\theta \]
6. \[ = a^2 \int_0^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta \]
7. আমরা জানি, \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) 😎
8. \[ = a^2 \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
9. \[ = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta \]
10. \[ = \frac{a^2}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}} \]
11. \[ = \frac{a^2}{2} \left[ \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right] \]
12. \[ = \frac{a^2}{2} \left[ \frac{\pi}{2} + 0 - 0 - 0 \right] \]
13. \[ = \frac{a^2}{2} \cdot \frac{\pi}{2} \]
14. \[ = \frac{a^2 \pi}{4} \]

অতএব, \( \int_0^a \sqrt{a^2 - x^2} \, dx = \frac{a^2 \pi}{4} \) 🎉