Question

Please give me the solution​

Answer 1

Answer:

$\large{\bold{\int\limits^1_0{x^\frac{5}{2}(1-x)^\frac{3}{2}}=\frac{3\pi}{32}}}$

Step-by-step explanation:

$I_{m,n}=\frac{m-1}{m+n}I_{m-2,n}\\\\m=6,\;\;n=4\;\\\\I_{6,4}=\frac{5}{10}I_{4,4}\\\\I_{6,4}=\frac{1}{2}I_{4,4}\\\\I_{6,4}=\frac{1}{2}\times\frac{3}{8}I_{2,4}\\\\I_{6,4}=\frac{1}{2}\times \frac{3}{8}\times \frac{1}{6}I_{0,4}\\\\\\I_{6,4}=\frac{1}{32}I_{0,4}$

$I=\int\limits^1_0 {x^\frac{5}{2}(1-x)^\frac{3}{2}} \, dx \\\\put\;x=sin^2\theta\\\\dx=2sin\theta cos\theta\\\\Limits\;\;x=0\;\;\;\;\theta=0\\Limits\;\;x=1\;\;\;\;\theta=\frac{\pi}{2}\\\\I=\int\limits^\frac{\pi}{2}_0{(sin^2\theta)^\frac{5}{2}(1-sin^2\theta)^\frac{3}{2}\times2sin\theta cos\theta}\, d\theta\\\\I=\int\limits^\frac{\pi}{2}_0{sin^5\theta(cos^2\theta)^\frac{3}{2}\times2sin\theta cos\theta}\, d\theta\\\\I=2\int\limits^\frac{\pi}{2}_0sin^6\theta cos^4\theta\,d\theta$

$I=\int\limits^\frac{\pi}{2}_0 {sin^4\theta} \, d\theta\\\\sin^2\theta=\frac{1-cos2\theta }{2}\\\\cos^2{2\theta}=\frac{1+cos4\theta}{2}\\\\I=\int\limits^\frac{\pi}{2}_0 (sin^2\theta)^2d\theta\\\\I=\int\limits^\frac{\pi}{2}_0(\frac{1-cos2\theta }{2})^2d\theta\\\\I=\frac{1}{4}\int\limits^\frac{\pi}{2}_0(1-2cos2\theta+cos^22\theta)d\theta\\\\I=\frac{1}{4}\int\limits^\frac{\pi}{2}_0(1-2cos2\theta+\frac{1+cos4\theta}{2})d\theta$

$I=\frac{1}{4}(\theta-sin2\theta+\frac{\theta}{2}+\frac{sin4\theta}{8})\limits^\frac{\pi}{2}_0\\\\I=\frac{1}{4}(\frac{3\pi}{8})\\\\I=\frac{3\pi}{32}$

HOPE IT HELPS