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Question Number 54765 by peter frank last updated on 10/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

(1/2)×2∫_0 ^(π/2) sin^(2×24−1) xcos^(2×27−1) xdx  formula 2∫_0 ^(π/2) sin^(2p−1) xcos^(2q−1) xdx  =((⌈(p)⌈(q))/(⌈(p+q)))  so ans is=(1/2)×((⌈(24)×⌈(27))/(⌈(51)))=(1/2)×((23!26!)/(50!))  firmula gamma function→⌈(n+1)=n!(factirial)

$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}×\mathrm{24}−\mathrm{1}} {xcos}^{\mathrm{2}×\mathrm{27}−\mathrm{1}} {xdx} \\ $$$${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{q}−\mathrm{1}} {xdx} \\ $$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$${so}\:{ans}\:{is}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\lceil\left(\mathrm{24}\right)×\lceil\left(\mathrm{27}\right)}{\lceil\left(\mathrm{51}\right)}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{23}!\mathrm{26}!}{\mathrm{50}!} \\ $$$${firmula}\:{gamma}\:{function}\rightarrow\lceil\left({n}+\mathrm{1}\right)={n}!\left({factirial}\right) \\ $$$$ \\ $$$$ \\ $$

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