int_0^(pi/2)cos^3x root.sinxdx  এর মান-

Explanation:

Another Explanation (5): সমাধান: ধরি, \( sinx = t^2 \) 🤩 সুতরাং, \( cosx dx = \frac{2tdt}{\sqrt{1-t^4}} \)🤔 অতএব, \( cosx = \sqrt{1-sin^2x} = \sqrt{1-t^4} \) 🤓 সুতরাং, \( \sqrt{1-t^4} dx = \frac{2tdt}{\sqrt{1-t^4}} \) বা \( dx = \frac{2tdt}{1-t^4} \) 🧐 এখন, যখন \( x = 0 \) তখন \( t = 0 \) এবং যখন \( x = \frac{\pi}{2} \) তখন \( t = 1 \) 😎 অতএব, \( \int_{0}^{\frac{\pi}{2}} cos^3x \sqrt{sinx} dx = \int_{0}^{1} (1-t^4)^{\frac{3}{2}} t \frac{2tdt}{1-t^4} \) ??? \( = 2\int_{0}^{1} (1-t^4)^{\frac{1}{2}} t^2 dt \) ✨ এখন, ধরি \( t^2 = u \) সুতরাং \( 2tdt = du \) 🙏 অতএব, \( \int_{0}^{1} (1-t^4)^{\frac{1}{2}} t^2 dt = \int_{0}^{1} (1-u^2)^{\frac{1}{2}} u^{\frac{1}{2}} \frac{1}{2}du \) 🥳 \( = \frac{1}{2}\int_{0}^{1} (1-u^2)^{\frac{1}{2}} u^{\frac{1}{2}} du \) 😇 আমরা জানি, \( \Beta(m, n) = 2\int_{0}^{1} x^{2m-1}(1-x^2)^{n-1} dx \) 🎈 এখানে, \( 2m-1 = \frac{1}{2} \) এবং \( n-1 = \frac{1}{2} \) 🎍 সুতরাং, \( m = \frac{3}{4} \) এবং \( n = \frac{3}{2} \) 🎊 অতএব, \( \int_{0}^{1} (1-u^2)^{\frac{1}{2}} u^{\frac{1}{2}} du = \frac{1}{2}\Beta(\frac{3}{4}, \frac{3}{2}) \) 🎎 আমরা জানি, \( \Beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \) 🎐 অতএব, \( \Beta(\frac{3}{4}, \frac{3}{2}) = \frac{\Gamma(\frac{3}{4})\Gamma(\frac{3}{2})}{\Gamma(\frac{9}{4})} \) 🎑 \( \Gamma(\frac{3}{2}) = \frac{1}{2}\Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{2} \) ۞ \( \Gamma(\frac{9}{4}) = \frac{5}{4}\Gamma(\frac{5}{4}) = \frac{5}{4}\cdot\frac{1}{4}\Gamma(\frac{1}{4}) = \frac{5}{16}\Gamma(\frac{1}{4}) \) 💝 সুতরাং, \( \frac{\Gamma(\frac{3}{4})\Gamma(\frac{3}{2})}{\Gamma(\frac{9}{4})} = \frac{\Gamma(\frac{3}{4})\frac{\sqrt{\pi}}{2}}{\frac{5}{16}\Gamma(\frac{1}{4})} = \frac{8\sqrt{\pi}}{5}\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \) 💘 আমরা জানি, \( \Gamma(x)\Gamma(1-x) = \frac{\pi}{sin(\pi x)} \) 💖 সুতরাং, \( \Gamma(\frac{1}{4})\Gamma(\frac{3}{4}) = \frac{\pi}{sin(\frac{\pi}{4})} = \pi\sqrt{2} \) 💞 অতএব, \( \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} = \frac{\pi\sqrt{2}}{\Gamma(\frac{1}{4})^2} \) 💓 সুতরাং, \( \frac{8\sqrt{\pi}}{5} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} = \frac{8\sqrt{\pi}}{5} \frac{\pi\sqrt{2}}{\Gamma(\frac{1}{4})^2} \) 💟 এখন, \( 2 \cdot \frac{1}{2} \Beta(\frac{3}{4}, \frac{3}{2}) = \Beta(\frac{3}{4}, \frac{3}{2}) = \frac{8}{5}\frac{\Gamma(\frac{3}{4})\Gamma(\frac{3}{2})}{\Gamma(\frac{9}{4})} \) 💜 \( = \frac{8}{5}\frac{\Gamma(\frac{3}{4})\frac{\sqrt{\pi}}{2}}{\frac{5}{4}\Gamma(\frac{5}{4})} \) 🫴 \( = \frac{8}{5} \cdot \frac{4}{5} \frac{\frac{\sqrt{\pi}}{2}\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})} \) 💔 \( = \frac{32}{25} \cdot \frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{3}{4})}{\frac{1}{4}\Gamma(\frac{1}{4})} \) 🫂 \( = \frac{32}{25} \cdot 2\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \) 😽 কিন্তু, \( \int_{0}^{\frac{\pi}{2}} cos^3x \sqrt{sinx} dx = \frac{8}{21} \) এটি সঠিক উত্তর। 😪 অন্যভাবে: \( \int_{0}^{\frac{\pi}{2}} cos^3x \sqrt{sinx} dx \) Let \( sinx = t^2 \), then \( cosx dx = 2tdt \) \( \int_{0}^{1} (1-t^4) \sqrt{t^2} 2tdt = \int_{0}^{1} 2t^2(1-t^4)dt \) \( = 2\int_{0}^{1} (t^2-t^6)dt = 2[\frac{t^3}{3} - \frac{t^7}{7}]_{0}^{1} \) \( = 2[\frac{1}{3}-\frac{1}{7}] = 2[\frac{7-3}{21}] = 2\cdot\frac{4}{21} = \frac{8}{21} \)