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Question Number 131401 by mnjuly1970 last updated on 04/Feb/21

$$\dots nicecalculus\dots$$$$provethat:::$$$${\int }_0^1\left(\mathcal{A}rcsin\left(x\right)\right).\left(\mathcal{A}rccos\left(x\right)\right)dx\stackrel{??}{=}2-\frac{\pi }{2}$$$$$$

Answered by mathmax by abdo last updated on 04/Feb/21

$$\mathrm{\Phi }={\int }_0^1\left(\mathrm{arcsinx}\right)\left(\mathrm{arcosx}\right)\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^1\mathrm{arcsinx}(\frac{\pi }{2}-\mathrm{arcsinx})\mathrm{dx}$$$$=\frac{\pi }{2}{\int }_0^1\mathrm{arcsinx}\mathrm{dx}-{\int }_0^1{\arcsin }^2\mathrm{x}\mathrm{dx}$$$${\int }_0^1\mathrm{arcsinx}\mathrm{dx}{=}_{\mathrm{arcsinx}=\mathrm{t}}{\int }_0^{\frac{\pi }{2}}\mathrm{t}.\mathrm{cost}\mathrm{dt}{=}_{\mathrm{byparts}}{\left[\mathrm{tsint}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\mathrm{sint}\mathrm{dt}$$$$=\frac{\pi }{2}+{\left[\mathrm{cost}\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-1$$$${\int }_0^1{\left(\mathrm{arcsinx}\right)}^2\mathrm{dx}{=}_{\mathrm{arcsinx}=\mathrm{t}}{\int }_0^{\frac{\pi }{2}}{\mathrm{t}}^2\mathrm{cost}\mathrm{dt}$$$$={\left[{\mathrm{t}}^2\mathrm{sint}\right]}_0^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}\mathrm{t}\mathrm{sint}\mathrm{dt}=\frac{{\pi }^2}{4}-2\{{[-\mathrm{tcost}]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\mathrm{cost}\mathrm{dt}\}$$$$=\frac{{\pi }^2}{4}-2{\left[\mathrm{sint}\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{4}-2\Rightarrow \mathrm{\Phi }=\frac{\pi }{2}(\frac{\pi }{2}-1)-\frac{{\pi }^2}{4}+2\Rightarrow$$$$\mathrm{\Phi }=2-\frac{\pi }{2}$$

Commented by mnjuly1970 last updated on 04/Feb/21

$$thankyousomuch$$$$sirmax\dots$$

Commented by mathmax by abdo last updated on 04/Feb/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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