Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 85872 by Wepa last updated on 25/Mar/20

$$\int {\cos }^{2020}\mathrm{x}\mathrm{dx}=?$$

Commented by mr W last updated on 25/Mar/20

$$I_n=\int {\cos }^{n}xdx$$$$=\int {\cos }^{n-1}xd\left(\sin x\right)$$$$=\sin x\cos {}^{n-1}x+(n-1)\int {\sin }^{2}x{\cos }^{n-2}xdx$$$$=\sin x\cos {}^{n-1}x+(n-1)\int (1-{\cos }^{2}x){\cos }^{n-2}xdx$$$$=\sin x\cos {}^{n-1}x+(n-1)\int {\cos }^{n-2}xdx-(n-1)\int {\cos }^{n}xdx$$$$=\sin x\cos {}^{n-1}x+(n-1)I_{n-2}-(n-1)I_n$$$${nI}_n=\sin x\cos {}^{n-1}x+(n-1)I_{n-2}$$$$\Rightarrow I_n=\frac{1}{n}\sin x\cos {}^{n-1}x+\frac{n-1}{n}{I}_{n-2}$$

Answered by mind is power last updated on 26/Mar/20

$$\int {cos}^n\left(x\right)dx$$$$-\int \frac{u^m}{\sqrt{1-u^2}}du$$$$\frac{1}{\sqrt{1-u^2}}=\sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}{u}^{2n+m}$$$$=-\int \frac{\left(2n\right)!u^{2n+m}}{2^{2n}{(n!)}^2}du$$$$=-\sum _{n⩾0}\frac{\left(2n\right)!u^{2n+m+1}}{2^{2n}{(n!)}^2(2n+m+1)}=-x^{m+1}\sum _{n⩾0}\frac{\underset{k=0}{\prod ^{n-1}}(k+\frac{1}{2})}{(2n+m+1)}.{\left(\frac{u^2}{2}\right)}^n$$$$=-x^{m+1}.\frac{2}{m+1}\sum _{k⩾0}\frac{\underset{k=0}{\prod ^{n-1}}(\frac{1}{2}+k).\underset{j=0}{\prod ^{n-1}}(\frac{m+1}{2}+j)}{2\underset{j=0}{\prod ^{n-1}}(j+\frac{m+3}{2})}.{\left(\frac{u^2}{2}\right)}^n.\frac{1}{n!}$$$$=\frac{-2u^{m+1}}{m+1}{.}_2{F}_1(\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{u^2}{2})+c$$$$\frac{-2{cos}^{m+1}\left(x\right)}{m+1}{}_2{F}_1(\frac{1}{2};\frac{m+1}{2};\frac{m+3}{2};\frac{{cos}^2\left(x\right)}{2})+c$$$$$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com