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Question Number 52898 by MJS last updated on 14/Jan/19

$$\int \arcsin x\arccos xdx=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$$a={sin}^{-}x$$$$sina=x$$$${sin}^{-1}x+{cos}^{-1}x=\frac{\pi }{2}$$$$\int a\times (\frac{\pi }{2}-a)\times cosada$$$$\frac{\pi }{2}\int acosada-\int a^{2}cosada$$$$\mathit{I}=\frac{\pi }{2}{I}_1-{\mathit{I}}_2$$$$I_1=\int acosada{I}_2=\int a^{2}cosada$$$$I_1=a\int cosada-\int [\frac{da}{da}\int cosada]da$$$$=asina-\int sinada$$$$=asina+cosa+c_1\leftarrow valueof{I}_1$$$$I_2=\int a^{2}cosada$$$$=a^{2}sina-\int 2asinada$$$$=a^{2}sina-I_3$$$$I_3=\int 2asinada$$$$=2a(-cosa)-\int 2\times -cosada$$$$=-2acosa+2sina$$$$\mathit{s}\mathit{o}{\mathit{I}}_2=a^{2}sina+2acosa-2sina$$$$\mathit{S}o\mathit{I}=\frac{\pi }{2}{\mathit{I}}_1-{\mathit{I}}_2$$$$=\frac{\pi }{2}(asina+cosa)-(a^{2}sina+2acosa-2sina)+c$$$$=\frac{\pi }{2}({xsin}^{-1}x+\sqrt{1-x^2})-[{\left({sin}^{-1}x\right)}^{2}x+2{sin}^{-1}x\times \sqrt{1-x^2}-2x)+c$$$$sirplscheck...$$$$$$$$$$

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