Question Number 177900 by yaslm last updated on 10/Oct/22
Answered by Ar Brandon last updated on 11/Oct/22
![I_n =∫_0 ^(π/2) sin^n xdx I_(n+2) =∫_0 ^(π/2) sin^(n+2) xdx =∫_0 ^(π/2) sin^n xsin^2 xdx=∫_0 ^(π/2) sin^n x(1−cos^2 x)dx =∫_0 ^(π/2) sin^n xdx−∫_0 ^(π/2) sin^n xcos^2 xdx =I_n −∫_0 ^(π/2) (sin^n xcosx)cosxdx =I_n −[((sin^(n+1) x)/(n+1))cosx]_0 ^(π/2) −(1/(n+1))∫_0 ^(π/2) (sin^(n+1) x)sinxdx =I_n −(1/(n+1))∫_0 ^(π/2) sin^(n+2) xdx=I_n −(1/(n+1))I_(n+2) ⇒(1+(1/(n+1)))I_(n+2) =I_n ⇒(((n+2)/(n+1)))I_(n+2) =I_n ⇒I_(n+2) =((n+1)/(n+2))I_n ⇒I_(2n) =((2n−1)/(2n))I_(2n−2) =((2n−1)/(2n))∙((2n−3)/(2n−2))∙((2n−5)/(2n−4))∙∙∙(3/4)∙(1/2)I_0 I_0 =∫_0 ^(π/2) (sinx)^0 dx=∫_0 ^(π/2) dx=(π/2) ⇒I_(2n) =((2n−1)/(2n))∙((2n−3)/(2n−2))∙((2n−5)/(2n−4))∙∙∙(3/4)∙(1/2)∙(π/2)](https://www.tinkutara.com/question/Q177903.png)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {xdx} \\ $$$${I}_{{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}+\mathrm{2}} {xdx} \\ $$$$\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {x}\mathrm{sin}^{\mathrm{2}} {xdx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {x}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {xdx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}} {x}\mathrm{cos}^{\mathrm{2}} {xdx} \\ $$$$\:\:\:\:\:\:\:\:\:={I}_{{n}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{{n}} {x}\mathrm{cos}{x}\right)\mathrm{cos}{xdx} \\ $$$$\:\:\:\:\:\:\:\:\:={I}_{{n}} −\left[\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}\mathrm{cos}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{{n}+\mathrm{1}} {x}\right)\mathrm{sin}{xdx} \\ $$$$\:\:\:\:\:\:\:\:\:={I}_{{n}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{n}+\mathrm{2}} {xdx}={I}_{{n}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}{I}_{{n}+\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){I}_{{n}+\mathrm{2}} ={I}_{{n}} \:\Rightarrow\left(\frac{{n}+\mathrm{2}}{{n}+\mathrm{1}}\right){I}_{{n}+\mathrm{2}} ={I}_{{n}} \:\Rightarrow{I}_{{n}+\mathrm{2}} =\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}{I}_{{n}} \\ $$$$\Rightarrow{I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{I}_{\mathrm{2}{n}−\mathrm{2}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}\centerdot\frac{\mathrm{2}{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{4}}\centerdot\centerdot\centerdot\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}{I}_{\mathrm{0}} \\ $$$${I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}{x}\right)^{\mathrm{0}} {dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}\centerdot\frac{\mathrm{2}{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{4}}\centerdot\centerdot\centerdot\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\pi}{\mathrm{2}} \\ $$
Commented by yaslm last updated on 11/Oct/22
thank you so much