int_0^4sqrt(16-x^2)dx = ? 

Explanation:

Another Explanation (5): ইন্টিগ্রেশনটির সমাধান নিচে দেওয়া হল: দেওয়া আছে, \(\int_0^4 \sqrt{16-x^2} \, dx\) ধরি, \(x = 4\sin(\theta)\) তাহলে, \(dx = 4\cos(\theta) \, d\theta\) যখন \(x = 0\), তখন \(4\sin(\theta) = 0 \implies \sin(\theta) = 0 \implies \theta = 0\) যখন \(x = 4\), তখন \(4\sin(\theta) = 4 \implies \sin(\theta) = 1 \implies \theta = \frac{\pi}{2}\) সুতরাং, ইন্টিগ্রালটি হবে: \(\int_0^{\frac{\pi}{2}} \sqrt{16 - (4\sin(\theta))^2} \cdot 4\cos(\theta) \, d\theta\) = \(\int_0^{\frac{\pi}{2}} \sqrt{16 - 16\sin^2(\theta)} \cdot 4\cos(\theta) \, d\theta\) = \(\int_0^{\frac{\pi}{2}} \sqrt{16(1 - \sin^2(\theta))} \cdot 4\cos(\theta) \, d\theta\) = \(\int_0^{\frac{\pi}{2}} 4\cos(\theta) \cdot 4\cos(\theta) \, d\theta\) = \(16 \int_0^{\frac{\pi}{2}} \cos^2(\theta) \, d\theta\) এখন, \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\) সুতরাং, ইন্টিগ্রালটি হবে: \(16 \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta\) = \(8 \int_0^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta\) = \(8 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}}\) = \(8 \left[ \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right]\) = \(8 \left[ \frac{\pi}{2} + 0 - 0 - 0 \right]\) = \(8 \cdot \frac{\pi}{2}\) = \(4\pi\) 🎉 সুতরাং, \(\int_0^4 \sqrt{16-x^2} \, dx = 4\pi\) ✨