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Question Number 163886 by zakirullah last updated on 11/Jan/22

Evaluate the following by using integration by parts formula: ∫xsin^(−1) (x)dx

$${Evaluate}\:{the}\:{following}\:{by}\:{using}\: \\ $$$$\:{integration}\:{by}\:{parts}\:{formula}: \\ $$$$\int{xsin}^{−\mathrm{1}} \left({x}\right){dx} \\ $$

Commented by Ar Brandon last updated on 11/Jan/22

u(x)=sin^(−1) (x) , v′(x)=x

$${u}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:\:,\:\:\:{v}'\left({x}\right)={x} \\ $$

Commented by zakirullah last updated on 11/Jan/22

dear Sir! need solution?

$${dear}\:\boldsymbol{{S}}{ir}!\:{need}\:{solution}? \\ $$

Answered by Ar Brandon last updated on 11/Jan/22

∫xsin^(−1) (x)dx=(1/2)x^2 sin^(−1) (x)−(1/2)∫(x^2 /( (√(1−x^2 ))))dx =(1/2)x^2 sin^(−1) (x)−(1/2)(x(√(1−x^2 ))−∫(√(1−x^2 ))dx) ∫(√(1−x^2 ))dx=(1/2)∫(1+cos2ϑ)dϑ , x=sinϑ =(1/2)(sin^(−1) (x)+x(√(1−x^2 )))+C ⇒∫xsin^(−1) (x)dx=(1/2)x^2 sin^(−1) (x)−(1/4)x(√(1−x^2 ))+(1/4)sin^(−1) (x)+C

$$\int{x}\mathrm{sin}^{−\mathrm{1}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\right) \\ $$$$\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos2}\vartheta\right){d}\vartheta\:,\:{x}=\mathrm{sin}\vartheta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+{C} \\ $$$$\Rightarrow\int{x}\mathrm{sin}^{−\mathrm{1}} \left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{4}}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{C} \\ $$

Commented by zakirullah last updated on 12/Jan/22

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