Step-by-step explanation:
[tex]\displaystyle \int_{0}^{2\pi}\cos^5x\:dx[/tex]
[tex]\displaystyle \int_{0}^{2\pi}\cos^5x\:dx[/tex]
[tex]\displaystyle \int_{0}^{2\pi}\cos x.\cos^4x\:dx[/tex]
[tex]\displaystyle \int_{0}^{2\pi}\cos x(\cos^2x)^2\:dx[/tex]
[tex]\because \: \cos^2x=1-\sin^2x[/tex]
[tex] x(1-\sin^2x)^2\:dx[/tex]
Now, we can write,
[tex]\displaystyle \int_{0}^{0}(1-t^2)^2\:dt[/tex]
As we know,
[tex]\displaystyle \int_{0}^{0}f(x)\:dx=0[/tex]
So,
[tex]\displaystyle \int_{0}^{0}(1-t^2)^2\:dt=0[/tex]
So, answer is Zero (0).
[tex]\displaystyle \int \sin x\:dx=-\cos x+C[/tex]
[tex] \sf{\displaystyle \int \cos x\:dx=\sin x+C}[/tex]
[tex] \sf{\displaystyle \int \tan x\:dx=ln|\sec x|+C}[/tex]