Integrate:
[tex] \ { \sf{ \pink{\displaystyle{ \int_0^{ \red{2\pi} cos^5 x\ dx∫02π​cos5x dx}}}}}[/tex]
Don't Answer If You Dont Know!








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Respuesta :

Step-by-step explanation:

Topic :-

  • Definite Integration

To Integrate :-

[tex]\displaystyle \int_{0}^{2\pi}\cos^5x\:dx[/tex]

Solution :-

[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]

we can write,

[tex] x(1-\sin^2x)^2\:dx[/tex]

  • Substitute sinx = t,

Differentiate both sides,

  • cosx.dx = dt

Change limits

Lower limit :-

  • sin(0) = 0

Upper limit :-

  • sin(2π) = 0

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]

Answer :-

So, answer is Zero (0).

Additional Formulae :-

[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]

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