মান নির্ণয় কর- 

int_0^(pi/2) (cos^3xsqrt(sinxdx))

প্রশ্ন: মান নির্ণয় কর- \( \int_0^{\frac{\pi}{2}} \cos^3x \sqrt{\sin x} \, dx \)

সমাধান:

আমরা জানি, \( \int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})} \)

এখানে, \( m = \frac{1}{2} \) এবং \( n = 3 \)

স??তরাং, \( \int_0^{\frac{\pi}{2}} \sin^{\frac{1}{2}} x \cos^3 x \, dx = \frac{\Gamma(\frac{\frac{1}{2}+1}{2}) \Gamma(\frac{3+1}{2})}{2\Gamma(\frac{\frac{1}{2}+3+2}{2})} \)

\(= \frac{\Gamma(\frac{3}{4}) \Gamma(2)}{2\Gamma(\frac{11}{4})} \)

আমরা জানি, \( \Gamma(x+1) = x\Gamma(x) \) এবং \( \Gamma(1) = 1 \), সুতরাং \( \Gamma(2) = 1! = 1 \)

\( \Gamma(\frac{11}{4}) = \Gamma(\frac{7}{4} + 1) = \frac{7}{4} \Gamma(\frac{7}{4}) \)

\( \Gamma(\frac{7}{4}) = \Gamma(\frac{3}{4} + 1) = \frac{3}{4} \Gamma(\frac{3}{4}) \)

সুতরাং, \( \Gamma(\frac{11}{4}) = \frac{7}{4} \cdot \frac{3}{4} \Gamma(\frac{3}{4}) = \frac{21}{16} \Gamma(\frac{3}{4}) \)

অতএব, \( \int_0^{\frac{\pi}{2}} \sin^{\frac{1}{2}} x \cos^3 x \, dx = \frac{\Gamma(\frac{3}{4}) \cdot 1}{2 \cdot \frac{21}{16} \Gamma(\frac{3}{4})} = \frac{1}{2 \cdot \frac{21}{16}} = \frac{16}{2 \cdot 21} = \frac{8}{21} \)

সুতরাং, \( \int_0^{\frac{\pi}{2}} \cos^3x \sqrt{\sin x} \, dx = \frac{8}{21} \) 🎉