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Question Number 85872 by Wepa last updated on 25/Mar/20
∫cos^(2020) x dx = ?
$$\int\mathrm{cos}^{\mathrm{2020}} \mathrm{x}\:\mathrm{dx}\:=\:? \\ $$
Commented by mr W last updated on 25/Mar/20
I_n =∫cos^n x dx =∫cos^(n−1) x d(sin x) =sin x cos^(n−1) x+(n−1)∫sin^2 x cos^(n−2) x dx =sin x cos^(n−1) x+(n−1)∫(1−cos^2 x) cos^(n−2) x dx =sin x cos^(n−1) x+(n−1)∫cos^(n−2) x dx−(n−1)∫cos^n x dx =sin x cos^(n−1) x+(n−1)I_(n−2) −(n−1)I_n nI_n =sin x cos^(n−1) x+(n−1)I_(n−2) ⇒I_n =(1/n)sin x cos^(n−1) x+((n−1)/n)I_(n−2)
$${I}_{{n}} =\int\mathrm{cos}^{{n}} \:{x}\:{dx} \\ $$$$=\int\mathrm{cos}^{{n}−\mathrm{1}} \:{x}\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\left({n}−\mathrm{1}\right)\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$=\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\left({n}−\mathrm{1}\right)\int\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}\right)\:\mathrm{cos}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$=\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\left({n}−\mathrm{1}\right)\int\mathrm{cos}^{{n}−\mathrm{2}} \:{x}\:{dx}−\left({n}−\mathrm{1}\right)\int\mathrm{cos}^{{n}} \:{x}\:{dx} \\ $$$$=\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} −\left({n}−\mathrm{1}\right){I}_{{n}} \\ $$$${nI}_{{n}} =\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{1}}{{n}}\mathrm{sin}\:{x}\:\mathrm{cos}\:^{{n}−\mathrm{1}} {x}+\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\ $$
Answered by mind is power last updated on 26/Mar/20
∫cos^n (x)dx −∫(u^m /( (√(1−u^2 ))))du (1/( (√(1−u^2 ))))=Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 ))u^(2n+m) =−∫(((2n)!u^(2n+m) )/(2^(2n) (n!)^2 ))du =−Σ_(n≥0) (((2n)!u^(2n+m+1) )/(2^(2n) (n!)^2 (2n+m+1)))=−x^(m+1) Σ_(n≥0) ((Π_(k=0) ^(n−1) (k+(1/2)))/((2n+m+1))).((u^2 /2))^n =−x^(m+1) .(2/(m+1))Σ_(k≥0) ((Π_(k=0) ^(n−1) ((1/2)+k).Π_(j=0) ^(n−1) (((m+1)/2)+j))/(2Π_(j=0) ^(n−1) (j+((m+3)/2)))).((u^2 /2))^n .(1/(n!)) =((−2u^(m+1) )/(m+1))._2 F_1 ((1/2),((m+1)/2);((m+3)/2);(u^2 /2))+c ((−2cos^(m+1) (x))/(m+1)) _2 F_1 ((1/2);((m+1)/2);((m+3)/2);((cos^2 (x))/2))+c
$$\int{cos}^{{n}} \left({x}\right){dx} \\ $$$$−\int\frac{{u}^{{m}} }{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{u}^{\mathrm{2}{n}+{m}} \\ $$$$=−\int\frac{\left(\mathrm{2}{n}\right)!{u}^{\mathrm{2}{n}+{m}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{du} \\ $$$$=−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{u}^{\mathrm{2}{n}+{m}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{2}{n}+{m}+\mathrm{1}\right)}=−{x}^{{m}+\mathrm{1}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+{m}+\mathrm{1}\right)}.\left(\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right)^{{n}} \\ $$$$=−{x}^{{m}+\mathrm{1}} .\frac{\mathrm{2}}{{m}+\mathrm{1}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).\underset{{j}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{{m}+\mathrm{1}}{\mathrm{2}}+{j}\right)}{\mathrm{2}\underset{{j}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({j}+\frac{{m}+\mathrm{3}}{\mathrm{2}}\right)}.\left(\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right)^{{n}} .\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{−\mathrm{2}{u}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}._{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{3}}{\mathrm{2}};\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right)+{c} \\ $$$$\frac{−\mathrm{2}{cos}^{{m}+\mathrm{1}} \left({x}\right)}{{m}+\mathrm{1}}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{3}}{\mathrm{2}};\frac{{cos}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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