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Question-202988




Question Number 202988 by sonukgindia last updated on 06/Jan/24
Commented by Frix last updated on 06/Jan/24
By parts.
$$\mathrm{By}\:\mathrm{parts}. \\ $$
Answered by MM42 last updated on 07/Jan/24
lnx=u⇒(dx/x)=du  sin^(−1) xdx=dv⇒xsin^(−1) x+(√(1−x^2 ))=v  ⇒I=xsin^(−1) xlnx+(√(1−x^2 ))lnx  −∫ sin^(−1) xdx−∫((√(1−x^2 ))/x)dx  ★ ∫ ((√(1−x^2 ))/x)dx→^(x=sinu) =∫ ((cos^2 u)/(sinu)) du  =∫ ((1/(sinu))−sinu)du=ln(tan(u/2))+cosu  =((1−(√(1−x^2 )))/x^2 )+(√(1−x^2 ))  ⇒I=(lnx−1)(xsin^(−1) x+(√(1−x^2 )))−((1+(√(1−x^2 )))/x^2 )−(√(1−x^2 ))   ✓
$${lnx}={u}\Rightarrow\frac{{dx}}{{x}}={du} \\ $$$${sin}^{−\mathrm{1}} {xdx}={dv}\Rightarrow{xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={v} \\ $$$$\Rightarrow{I}={xsin}^{−\mathrm{1}} {xlnx}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{lnx} \\ $$$$−\int\:{sin}^{−\mathrm{1}} {xdx}−\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}{dx} \\ $$$$\bigstar\:\int\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}{dx}\overset{{x}={sinu}} {\rightarrow}=\int\:\frac{{cos}^{\mathrm{2}} {u}}{{sinu}}\:{du} \\ $$$$=\int\:\left(\frac{\mathrm{1}}{{sinu}}−{sinu}\right){du}={ln}\left({tan}\frac{{u}}{\mathrm{2}}\right)+{cosu} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=\left({lnx}−\mathrm{1}\right)\left({xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:\checkmark \\ $$$$ \\ $$
Commented by MathematicalUser2357 last updated on 11/Jan/24
(ln x−1)(xsin^(−1) x+(√(1−x^2 )))−((1+(√(1−x^2 )))/x^2 )−(√(1−x^2 ))+C
$$\left(\mathrm{ln}\:{x}−\mathrm{1}\right)\left({x}\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{C} \\ $$

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