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If $\displaystyle I_{1}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{12}dx$ and $\displaystyle I_{2}=\int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ and $I_{1}=k(108\sqrt{3})I_{2}$. Then $k$ is

Try: put $x=\sin^2\theta$. Then $dx=2\sin\theta \cos\theta d\theta$.

so $$I_{1}=\frac{1}{6}\int^{\frac{\pi}{2}}_{0}\sin^6\theta \cos ^{8}\theta d\theta=\frac{5\cdot 3\cdot 7\cdot 5 \cdot 3}{14\cdot 12\cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2}\times \frac{\pi}{2}$$

Could some helo me to solve second one , Thanks

DXT
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  • For $\displaystyle I_{2},$ Put $\displaystyle x=\frac{1}{t}$ and Then $(3t+1)=u$ and Then $\displaystyle u=4\sec^2(\theta)$ – DXT Apr 29 '23 at 01:35

1 Answers1

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Just try to convert $I_1$ into $I_2$ form

Substituting $x = \cfrac{4y}{3+y}$

$$I_1 =\int_0^1 \cfrac{4^{\cfrac{5}2}3^{\cfrac72} x^{\cfrac52}(1-x)^{\cfrac72}}{(3+x)^6} \cfrac{4(3+x)-4x}{(3+x)^2}dx\\ \hspace{-1cm}= 32*27*\sqrt 3 *12*I_1 \\ \hspace{-1cm} =10368\sqrt 3*I_1 \\ \hspace{0cm}= 96*108*\sqrt 3 I_1$$

$\implies k = 96$

kayush
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