Find out ∫ecos⁡xdx

sunshine022uv

sunshine022uv

Answered

2022-01-03

Find out
ecosxdx

Answer & Explanation

Robert Pina

Robert Pina

Expert

2022-01-04Added 42 answers

First of all: there is no close form solution in terms of elementary functions.
What can you do, but its
Anzante2m

Anzante2m

Expert

2022-01-05Added 34 answers

This is just a response to your comments about having tried integration by parts for this problem.
Youre
karton

karton

Expert

2022-01-11Added 439 answers

ecosxdx=n=0cos2nx(2n)!dx+n=0cos2n+1x(2n+1)!dx=(1+n=1cos2nx(2n)!)dx+n=0cos2n+1x(2n+1)!dxFor n is any natural number,cos2nxdx=(2n)!x4n(n!)2+k=1n(2n)!((k1)!)2sinxcos2k1x4nk+1(n!)2(2k1)!+CThis result can be done by successive integration by parts.For n is any non-negative integer,cos2n+1xdx=cos2nxd(sinx)=(1sin2x)nd(sinx)=k=0Ckn(1)ksin2kxd(sinx)=k=0n(1)kn!sin2k+1xk!(nk)!(2k+1)+C(1+n=1cos2nx(2n)!)dx+n=0cos2n+1x(2n+1)!dx=x+n=1x4n(n!)2+n=1k=1n(k1)!)2sinxcos2k1x4nk+1(n!)2(2k1)!+k=0n(1)kn!sin2k+1x(2n+1)!k!(nk)!(2k+1)+C

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