Question Number 115014 by mathdave last updated on 23/Sep/20
$${if}\:{I}_{{n}} =\int_{{x}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{cos}^{{n}} {xdx},{where}\:{n}\succ\mathrm{1}\:{show} \\ $$$${that}\:{I}_{{n}} =\frac{{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} −\mathrm{1}}{{n}^{\mathrm{2}} }\:{and}\:{then} \\ $$$${evaluate}\:\:\int_{{x}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{cos}^{\mathrm{8}} {xdx} \\ $$
Answered by Olaf last updated on 23/Sep/20
![I_n = ∫_0 ^(π/2) xcos^n xdx I_n −I_(n+2) = ∫_0 ^(π/2) xcos^n x(1−cos^2 x)dx I_n −I_(n+2) = −∫_0 ^(π/2) (xsinx)(−sinxcos^n x)dx I_n −I_(n+2) = −[xsinx((cos^(n+1) x)/(n+1))]_0 ^(π/2) +∫_0 ^(π/2) (sinx+xcosx)((cos^(n+1) x)/(n+1))dx I_n −I_(n+2) = −∫_0 ^(π/2) (−sinx((cos^(n+1) x)/(n+1)))dx +(1/(n+1))I_(n+2) I_n −I_(n+2) = −[((cos^(n+2) x)/((n+1)(n+2)))]_0 ^(π/2) +(1/(n+1))I_(n+2) I_n = (1/((n+1)(n+2)))+I_(n+2) +(1/(n+1))I_(n+2) I_n = ((n+2)/(n+1))I_(n+2) +(1/((n+1)(n+2))) I_(n+2) = ((n+1)/(n+2))I_n −(1/((n+2)^2 )) (1) I_n = ((n(n−1)I_(n−2) −1)/n^2 ) I_0 = ∫_0 ^(π/2) xdx = [(x^2 /2)]_0 ^(π/2) = (π^2 /8) With (1) : I_2 = (1/2)((π^2 /8))−(1/4) = (π^2 /(16))−(1/4) I_4 = (3/4)((π^2 /(16))−(1/4))−(1/(16)) = ((3π^2 )/(64))−(1/4) I_6 = (5/6)(((3π^2 )/(64))−(1/4))−(1/(36)) = ((5π^2 )/(128))−((17)/(72)) I_8 = (7/8)(((5π^2 )/(128))−((17)/(72)))−(1/(64)) = ((35π^2 )/(1024))−(2/9) I_8 = ∫_0 ^(π/2) xcos^8 xdx = ((35π^2 )/(1024))−(2/9)](https://www.tinkutara.com/question/Q115155.png)
$$ \\ $$$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{cos}^{{n}} {xdx} \\ $$$$\mathrm{I}_{{n}} −\mathrm{I}_{{n}+\mathrm{2}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{cos}^{{n}} {x}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right){dx} \\ $$$$\mathrm{I}_{{n}} −\mathrm{I}_{{n}+\mathrm{2}} \:=\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({x}\mathrm{sin}{x}\right)\left(−\mathrm{sin}{x}\mathrm{cos}^{{n}} {x}\right){dx} \\ $$$$\mathrm{I}_{{n}} −\mathrm{I}_{{n}+\mathrm{2}} \:=\:−\left[{x}\mathrm{sin}{x}\frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}{x}+{x}\mathrm{cos}{x}\right)\frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}{dx} \\ $$$$\mathrm{I}_{{n}} −\mathrm{I}_{{n}+\mathrm{2}} \:=\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−\mathrm{sin}{x}\frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}\right){dx} \\ $$$$+\frac{\mathrm{1}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} \\ $$$$\mathrm{I}_{{n}} −\mathrm{I}_{{n}+\mathrm{2}} \:=\:−\left[\frac{\mathrm{cos}^{{n}+\mathrm{2}} {x}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} \\ $$$$\mathrm{I}_{{n}} \:=\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}+\mathrm{I}_{{n}+\mathrm{2}} \:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} \\ $$$$\mathrm{I}_{{n}} \:=\:\frac{{n}+\mathrm{2}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} +\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\mathrm{I}_{{n}+\mathrm{2}} \:=\:\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\mathrm{I}_{{n}} −\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }\:\left(\mathrm{1}\right) \\ $$$$\mathrm{I}_{{n}} \:=\:\frac{{n}\left({n}−\mathrm{1}\right)\mathrm{I}_{{n}−\mathrm{2}} −\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xdx}\:=\:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{With}\:\left(\mathrm{1}\right)\::\:\mathrm{I}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{I}_{\mathrm{4}} \:=\:\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{16}}\:=\:\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{64}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{I}_{\mathrm{6}} \:=\:\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{64}}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{36}}\:=\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{128}}−\frac{\mathrm{17}}{\mathrm{72}} \\ $$$$\mathrm{I}_{\mathrm{8}} \:=\:\frac{\mathrm{7}}{\mathrm{8}}\left(\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{128}}−\frac{\mathrm{17}}{\mathrm{72}}\right)−\frac{\mathrm{1}}{\mathrm{64}}\:=\:\frac{\mathrm{35}\pi^{\mathrm{2}} }{\mathrm{1024}}−\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{I}_{\mathrm{8}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\mathrm{cos}^{\mathrm{8}} {xdx}\:=\:\frac{\mathrm{35}\pi^{\mathrm{2}} }{\mathrm{1024}}−\frac{\mathrm{2}}{\mathrm{9}} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$