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

∫arcsin x arccos x dx=?

$$\int\mathrm{arcsin}\:{x}\:\mathrm{arccos}\:{x}\:{dx}=? \\ $$

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

a=sin^− x  sina=x  sin^(−1) x+cos^(−1) x=(π/2)  ∫a×((π/2)−a)×cosada  (π/2)∫acosada−∫a^2 cosada  I=(π/2)I_1 −I_2   I_1 =∫acosada   I_2 =∫a^2 cosada  I_1 =a∫cosada−∫[(da/da)∫cosada]da  =asina−∫sinada  =asina+cosa+c_1   ←value of I_1   I_2 =∫a^2 cosada  =a^2 sina−∫2asinada  =a^2 sina−I_3   I_3 =∫2asinada  =2a(−cosa)−∫2×−cosada  =−2acosa+2sina  so I_2 =a^2 sina+2acosa−2sina  So I=(π/2)I_1 −I_2   =(π/2)(asina+cosa)−(a^2 sina+2acosa−2sina)+c  =(π/2)(xsin^(−1) x+(√(1−x^2 )) )−[(sin^(−1) x)^2 x+2sin^(−1) x×(√(1−x^2 )) −2x)+c  sir pls check...

$${a}={sin}^{−} {x} \\ $$$${sina}={x} \\ $$$${sin}^{−\mathrm{1}} {x}+{cos}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{2}} \\ $$$$\int{a}×\left(\frac{\pi}{\mathrm{2}}−{a}\right)×{cosada} \\ $$$$\frac{\pi}{\mathrm{2}}\int{acosada}−\int{a}^{\mathrm{2}} {cosada} \\ $$$$\boldsymbol{{I}}=\frac{\pi}{\mathrm{2}}{I}_{\mathrm{1}} −\boldsymbol{{I}}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int{acosada}\:\:\:{I}_{\mathrm{2}} =\int{a}^{\mathrm{2}} {cosada} \\ $$$${I}_{\mathrm{1}} ={a}\int{cosada}−\int\left[\frac{{da}}{{da}}\int{cosada}\right]{da} \\ $$$$={asina}−\int{sinada} \\ $$$$={asina}+{cosa}+{c}_{\mathrm{1}} \:\:\leftarrow{value}\:{of}\:{I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int{a}^{\mathrm{2}} {cosada} \\ $$$$={a}^{\mathrm{2}} {sina}−\int\mathrm{2}{asinada} \\ $$$$={a}^{\mathrm{2}} {sina}−{I}_{\mathrm{3}} \\ $$$${I}_{\mathrm{3}} =\int\mathrm{2}{asinada} \\ $$$$=\mathrm{2}{a}\left(−{cosa}\right)−\int\mathrm{2}×−{cosada} \\ $$$$=−\mathrm{2}{acosa}+\mathrm{2}{sina} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{I}}_{\mathrm{2}} ={a}^{\mathrm{2}} {sina}+\mathrm{2}{acosa}−\mathrm{2}{sina} \\ $$$$\boldsymbol{{S}}{o}\:\boldsymbol{{I}}=\frac{\pi}{\mathrm{2}}\boldsymbol{{I}}_{\mathrm{1}} −\boldsymbol{{I}}_{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left({asina}+{cosa}\right)−\left({a}^{\mathrm{2}} {sina}+\mathrm{2}{acosa}−\mathrm{2}{sina}\right)+{c} \\ $$$$=\frac{\pi}{\mathrm{2}}\left({xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\right)−\left[\left({sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} {x}+\mathrm{2}{sin}^{−\mathrm{1}} {x}×\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:−\mathrm{2}{x}\right)+{c} \\ $$$${sir}\:{pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$

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