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Question Number 226952 by Spillover last updated on 20/Dec/25
Answered by Spillover last updated on 24/Dec/25
(a) I_n =∫_0 ^(π/2) e^(−x) cos^n xdx u=cos^n x ∗spillover∗ du=−nsin xcos^(n−1) xdx dv=e^(−x) dx v=−e^(−x) ∫udv=uv−∫vdu uv=[e^(−x) cos^n x]_0 ^(π/2) =1 ∫vdu=∫_0 ^(π/2) −e^(−x) −nsin xcos^(n−1) x ∫vdu=n∫_0 ^(π/2) e^(−x) sin xcos^(n−1) xdx hence ∗spillover∗ 1−n∫_0 ^(π/2) e^(−x) sin xcos^(n−1) xdx
$$\left({a}\right) \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:^{{n}} {xdx} \\ $$$${u}=\mathrm{cos}\:^{{n}} {x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast{spillover}\ast \\ $$$${du}=−{n}\mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {xdx} \\ $$$${dv}={e}^{−{x}} {dx}\:\:\:\:{v}=−{e}^{−{x}} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$${uv}=\left[{e}^{−{x}} \mathrm{cos}\:^{{n}} {x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1} \\ $$$$\int{vdu}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −{e}^{−{x}} −{n}\mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {x} \\ $$$$\int{vdu}={n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {xdx} \\ $$$${hence}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast{spillover}\ast \\ $$$$\mathrm{1}−{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {xdx} \\ $$$$ \\ $$
Answered by Spillover last updated on 24/Dec/25
(b) (n^2 +1)I_n =1+n(n−1)I_(n−2) ∗spillover∗ u=sin xcos^(n−1) x v=−e^(−x) du=cos xcos^(n−1) x−(n−1)cos^(n−2) xsin^2 xdx du=cos xcos^(n−1) x−(n−1)cos^(n−2) x(1−sin^2 x)dx du=ncos^n x−(n−1)cos^(n−2) xdx From I_n =1−n∫_0 ^(π/2) e^(−x) sin xcos^(n−1) xdx ∫udu=uv−∫vdu uv=[e^(−x) sin xcos^(n−1) x]_0 ^(π/2) =0 ∫vdu v=−e^(−x) du=ncos^n x−(n−1)cos^(n−2) xdx ∫(−e^(−x) )ncos^n x−(n−1)cos^(n−2) xdx n∫e^(−x) cos^n xdx−(n−1)∫_0 ^(π/2) e^(−x) cos^(n−1) xdx ∗spillover∗ but ∫e^(−x) cos^n xdx=I_n ∫e^(−x) cos^(n−2) xdx=I_(n−2) I_n =1−n(nI_n −(n−1) I_n =1−n(nI_n −(n−1)I_(n−2) I_n +n^2 I_n =1+n(n−1)uu (n^2 +1)I_n =1+n(n−1)I_(n−2)
$$\left({b}\right) \\ $$$$ \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \:\:\:\:\:\ast{spillover}\ast \\ $$$${u}=\mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {x}\:\:\:\:\:{v}=−{e}^{−{x}} \\ $$$${du}=\mathrm{cos}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {x}−\left({n}−\mathrm{1}\right)\mathrm{cos}\:^{{n}−\mathrm{2}} {x}\mathrm{sin}\:^{\mathrm{2}} {xdx} \\ $$$${du}=\mathrm{cos}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {x}−\left({n}−\mathrm{1}\right)\mathrm{cos}\:^{{n}−\mathrm{2}} {x}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right){dx} \\ $$$${du}={n}\mathrm{cos}\:^{{n}} {x}−\left({n}−\mathrm{1}\right)\mathrm{cos}\:^{{n}−\mathrm{2}} {xdx} \\ $$$${From} \\ $$$${I}_{{n}} =\mathrm{1}−{n}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {xdx} \\ $$$$\int{udu}={uv}−\int{vdu} \\ $$$${uv}=\left[{e}^{−{x}} \mathrm{sin}\:{x}\mathrm{cos}\:^{{n}−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{0} \\ $$$$\int{vdu} \\ $$$${v}=−{e}^{−{x}} \\ $$$${du}={n}\mathrm{cos}\:^{{n}} {x}−\left({n}−\mathrm{1}\right)\mathrm{cos}\:^{{n}−\mathrm{2}} {xdx} \\ $$$$\int\left(−{e}^{−{x}} \right){n}\mathrm{cos}\:^{{n}} {x}−\left({n}−\mathrm{1}\right)\mathrm{cos}\:^{{n}−\mathrm{2}} {xdx} \\ $$$${n}\int{e}^{−{x}} \mathrm{cos}\:^{{n}} {xdx}−\left({n}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:^{{n}−\mathrm{1}} {xdx}\:\:\:\:\:\:\ast{spillover}\ast \\ $$$${but}\:\:\int{e}^{−{x}} \mathrm{cos}\:^{{n}} {xdx}={I}_{{n}} \\ $$$$\:\:\:\:\:\:\:\:\int{e}^{−{x}} \mathrm{cos}\:^{{n}−\mathrm{2}} {xdx}={I}_{{n}−\mathrm{2}} \\ $$$${I}_{{n}} =\mathrm{1}−{n}\left({nI}_{{n}} −\left({n}−\mathrm{1}\right)\right. \\ $$$${I}_{{n}} =\mathrm{1}−{n}\left({nI}_{{n}} −\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \right. \\ $$$${I}_{{n}} +{n}^{\mathrm{2}} {I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){uu} \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$$ \\ $$
Answered by Spillover last updated on 24/Dec/25
(c) I_6 =((263−144e^(−(π/2)) )/(629)) from (n^2 +1)I_n =1+n(n−1)I_(n−1) n=2 5I_2 =1+2I_0 I_0 =? from I_n =∫_0 ^(π/2) e^(−x) cos^(n−1) xdx I_o =∫_0 ^(π/2) e^(−x) cos^(1−1) xdx=∫_0 ^(π/2) e^(−x) dx I_1 =∫_0 ^(π/2) e^(−x) dx=[−e^(−x) ]_0 ^(π/2) =1−e^(−(π/2)) I_1 =∫_0 ^(π/2) e^(−x) cos xdx u=cos x du=−sin xdx dv=e^(−x) dx v=−e^(−x) ∫udv=uv−∫vdu [−e^(−x) cos x]_0 ^(π/2) −∫_0 ^(π/2) e^(−x) sin xdx=1+∫_0 ^(π/2) e^(−x) sin xdx I_n =∫_0 ^(π/2) e^(−x) sin xdx ∗ spillover∗ I_1 =∫_0 ^(π/2) e^(−x) sin xdx u=sin x du=cos xdx dv=e^(−x) dx v=e^(−x) ∫udv=uv−∫vdu [−e^(−x) sin x]_0 ^(π/2) −∫_0 ^(π/2) (−e^(−x) )cos xdx I_1 =−e^(−(π/2)) −0+∫_0 ^(π/2) e^(−x) cos xdx=−e^(−(π/2)) +I_1 I_1 =1−(−e^(−(π/2)) +I_1 )=I_1 =((1+e^(−(π/2)) )/2) page 1
$$\left({c}\right) \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{629}} \\ $$$${from} \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{1}} \\ $$$${n}=\mathrm{2} \\ $$$$\mathrm{5}{I}_{\mathrm{2}} =\mathrm{1}+\mathrm{2}{I}_{\mathrm{0}} \:\:\:\:\:\:\:{I}_{\mathrm{0}} =? \\ $$$${from} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:^{{n}−\mathrm{1}} {xdx} \\ $$$${I}_{{o}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:^{\mathrm{1}−\mathrm{1}} {xdx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} {dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} {dx}=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1}−{e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:{xdx}\:\:\:\:\:\:\:\:\:{u}=\mathrm{cos}\:{x}\:\:\:{du}=−\mathrm{sin}\:{xdx} \\ $$$${dv}={e}^{−{x}} {dx}\:\:\:\:\:{v}=−{e}^{−{x}} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$$\left[−{e}^{−{x}} \mathrm{cos}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{xdx}=\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{xdx} \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{xdx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\:{spillover}\ast \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{sin}\:{xdx} \\ $$$${u}=\mathrm{sin}\:{x}\:\:\:\:\:\:\:{du}=\mathrm{cos}\:{xdx}\:\:\:\:{dv}={e}^{−{x}} {dx}\:\:\:\:\:\:\:{v}={e}^{−{x}} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$$\left[−{e}^{−{x}} \mathrm{sin}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−{e}^{−{x}} \right)\mathrm{cos}\:{xdx} \\ $$$${I}_{\mathrm{1}} =−{e}^{−\frac{\pi}{\mathrm{2}}} −\mathrm{0}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{−{x}} \mathrm{cos}\:{xdx}=−{e}^{−\frac{\pi}{\mathrm{2}}} +{I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\mathrm{1}−\left(−{e}^{−\frac{\pi}{\mathrm{2}}} +{I}_{\mathrm{1}} \right)={I}_{\mathrm{1}} =\frac{\mathrm{1}+{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{2}} \\ $$$${page}\:\mathrm{1} \\ $$$$ \\ $$
Answered by Spillover last updated on 24/Dec/25
from (n^2 +1)I_n =1+n(n−1)I_(n−2) n=2 5I_2 =1−2I_0 but I_0 =1−e^(π/2) 5I_2 =1−2I_0 ∗spillover∗ 5I_2 =1+2(1−e^(π/2) )=((3−2e^(π/2) )/5) I_2 =((3−2e^(π/2) )/5) for n=4 (n^2 +1)I_n =1+n(n−1)I_(n−2) 17I_4 =1+12I_2 17I_4 =1+12(((3−2e^(π/2) )/5)) 17I_4 =41−24e^(π/2) I_4 =((41−24e^(π/2) )/(17)) for n=6 ∗spillover∗ (n^2 +1)I_n =1+n(n−1)I_(n−2) 37I_6 =1+30I_4 37I_6 =1+30(((41−24e^(π/2) )/(17)) )=((263−144e^(−(π/2)) )/(17)) 37I_6 =((263−144e^(−(π/2)) )/(17)) I_6 =((263−144e^(−(π/2)) )/(17×37))=((263−144e^(−(π/2)) )/(629)) I_6 =((263−144e^(−(π/2)) )/(629)) page 2
$${from}\:\:\:\:\:\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$${n}=\mathrm{2}\:\:\:\:\:\:\:\mathrm{5}{I}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}{I}_{\mathrm{0}} \:\:\:\:{but}\:{I}_{\mathrm{0}} =\mathrm{1}−{e}^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{5}{I}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}{I}_{\mathrm{0}} \:\:\:\:\:\:\:\:\:\ast{spillover}\ast \\ $$$$\mathrm{5}{I}_{\mathrm{2}} =\mathrm{1}+\mathrm{2}\left(\mathrm{1}−{e}^{\frac{\pi}{\mathrm{2}}} \right)=\frac{\mathrm{3}−\mathrm{2}{e}^{\frac{\pi}{\mathrm{2}}} }{\mathrm{5}}\:\:\:\:\:\:\:\:{I}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{2}{e}^{\frac{\pi}{\mathrm{2}}} }{\mathrm{5}} \\ $$$${for}\:{n}=\mathrm{4} \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$$\mathrm{17}{I}_{\mathrm{4}} =\mathrm{1}+\mathrm{12}{I}_{\mathrm{2}} \:\:\:\:\:\:\:\mathrm{17}{I}_{\mathrm{4}} =\mathrm{1}+\mathrm{12}\left(\frac{\mathrm{3}−\mathrm{2}{e}^{\frac{\pi}{\mathrm{2}}} }{\mathrm{5}}\right) \\ $$$$\mathrm{17}{I}_{\mathrm{4}} =\mathrm{41}−\mathrm{24}{e}^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\:\:\:\:{I}_{\mathrm{4}} =\frac{\mathrm{41}−\mathrm{24}{e}^{\frac{\pi}{\mathrm{2}}} \:}{\mathrm{17}}\: \\ $$$${for}\:{n}=\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast{spillover}\ast \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} =\mathrm{1}+{n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$$\mathrm{37}{I}_{\mathrm{6}} =\mathrm{1}+\mathrm{30}{I}_{\mathrm{4}} \:\:\:\:\:\:\:\:\:\:\:\mathrm{37}{I}_{\mathrm{6}} =\mathrm{1}+\mathrm{30}\left(\frac{\mathrm{41}−\mathrm{24}{e}^{\frac{\pi}{\mathrm{2}}} \:}{\mathrm{17}}\:\right)=\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{17}} \\ $$$$\mathrm{37}{I}_{\mathrm{6}} =\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{17}} \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{17}×\mathrm{37}}=\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{629}}\: \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{263}−\mathrm{144}{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{629}}\: \\ $$$${page}\:\mathrm{2} \\ $$

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