Question Number 159823 by tounghoungko last updated on 21/Nov/21
Answered by Ar Brandon last updated on 21/Nov/21
![I=∫_0 ^1 ((√(1−x))/( (√(1+x))))sin^(−1) xdx=∫_0 ^1 ((1−x)/( (√(1−x^2 ))))sin^(−1) xdx =∫_0 ^1 ((sin^(−1) x)/( (√(1−x^2 ))))dx−∫_0 ^1 (x/( (√(1−x^2 ))))sin^(−1) xdx =[(((sin^(−1) x)^2 )/2)]_0 ^1 −[−(√(1−x^2 ))sin^(−1) x+∫((√(1−x^2 ))/( (√(1−x^2 ))))dx]_0 ^1 =(π^2 /8)−∫_0 ^1 dx=(π^2 /8)−1](https://www.tinkutara.com/question/Q159826.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}}\mathrm{sin}^{−\mathrm{1}} {xdx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mathrm{sin}^{−\mathrm{1}} {xdx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mathrm{sin}^{−\mathrm{1}} {xdx} \\ $$$$\:\:\:=\left[\frac{\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\left[−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\mathrm{sin}^{−\mathrm{1}} {x}+\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\mathrm{1}} {dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\mathrm{1} \\ $$
Commented by Ar Brandon last updated on 21/Nov/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}.\:\mathrm{Let}\:{s}=\mathrm{sin}^{−\mathrm{1}} {x}\Rightarrow{ds}=\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mathrm{sin}^{−\mathrm{1}} {xdx} \\ $$$$\mathrm{By}\:\mathrm{part}\:\begin{cases}{{u}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}} {x}}\\{\mathrm{v}'\left({x}\right)=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}\end{cases} \\ $$
Commented by tounghoungko last updated on 21/Nov/21
$${awesome} \\ $$
Answered by Kunal12588 last updated on 21/Nov/21
![∫_0 ^1 (((√(1−x)) sin^(−1) x)/( (√(1+x))))dx I=∫(((√(1−x)) sin^(−1) x)/( (√(1+x))))dx u=sin^(−1) x , dv=(√((1−x)/(1+x)))dx ⇒du=(1/( (√(1−x^2 ))))dx v=∫(√((1−x)/(1+x)))dx [taking x = sin u] v=∫(√((1−sin u)/(1+sin u)))×cos u du=∫((1−sin u)/(cos u))×cos u du ⇒v = u+cos u = sin^(−1) x+(√(1−x^2 )) I=sin^(−1) x(sin^(−1) x+(√(1−x^2 )))−∫(( sin^(−1) x+(√(1−x^2 )))/( (√(1−x^2 ))))dx I=sin^(−1) x(sin^(−1) x+(√(1−x^2 )))−∫((sin^(−1) x)/( (√(1−x^2 ))))dx−∫dx let t=sin^(−1) x⇒dt=(1/( (√(1−x^2 ))))dx I=sin^(−1) x(sin^(−1) x+(√(1−x^2 )))−∫tdt−x I=sin^(−1) x(sin^(−1) x+(√(1−x^2 )))−x−(1/2)t^2 I=sin^(−1) x(sin^(−1) x+(√(1−x^2 )))−x−(1/2)(sin^(−1) x)^2 +C ⇒I=(1/2)(sin^(−1) x)^2 +(√(1−x^2 ))sin^(−1) x −x+C ∫_0 ^1 (((√(1−x)) sin^(−1) x)/( (√(1+x))))dx =[(1/2)((π/2))^2 +0×(π/2)−1]−[(1/2)×0+0−0] ∫_0 ^1 (((√(1−x)) sin^(−1) x)/( (√(1+x))))dx=(π^2 /8)−1](https://www.tinkutara.com/question/Q159838.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}+{x}}}{dx} \\ $$$${I}=\int\frac{\sqrt{\mathrm{1}−{x}}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}+{x}}}{dx} \\ $$$${u}=\mathrm{sin}^{−\mathrm{1}} {x}\:,\:\:{dv}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx} \\ $$$$\Rightarrow{du}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${v}=\int\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\:\:\:\:\left[{taking}\:{x}\:=\:\mathrm{sin}\:{u}\right] \\ $$$${v}=\int\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:{u}}{\mathrm{1}+\mathrm{sin}\:{u}}}×\mathrm{cos}\:{u}\:{du}=\int\frac{\mathrm{1}−\mathrm{sin}\:{u}}{\mathrm{cos}\:{u}}×\mathrm{cos}\:{u}\:{du} \\ $$$$\Rightarrow{v}\:=\:{u}+\mathrm{cos}\:{u}\:=\:\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}\left(\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int\frac{\:\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}\left(\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int{dx} \\ $$$${let}\:{t}=\mathrm{sin}^{−\mathrm{1}} {x}\Rightarrow{dt}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}\left(\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\int{tdt}−{x} \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}\left(\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−{x}−\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$${I}=\mathrm{sin}^{−\mathrm{1}} {x}\left(\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−{x}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} +{C} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} +\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\mathrm{sin}^{−\mathrm{1}} {x}\:−{x}+{C} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}+{x}}}{dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{0}×\frac{\pi}{\mathrm{2}}−\mathrm{1}\right]−\left[\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}+\mathrm{0}−\mathrm{0}\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{1}−{x}}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}+{x}}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\mathrm{1} \\ $$
Commented by tounghoungko last updated on 21/Nov/21
$${yess} \\ $$