Question

Integrate:

Answer 1

Step-by-step explanation:

Topic :-

• Definite Integration

To Integrate :-

$\displaystyle \int_{0}^{2\pi}\cos^5x\:dx$

Solution :-

$\displaystyle \int_{0}^{2\pi}\cos^5x\:dx$

$\displaystyle \int_{0}^{2\pi}\cos x.\cos^4x\:dx$

$\displaystyle \int_{0}^{2\pi}\cos x(\cos^2x)^2\:dx$

$\because \: \cos^2x=1-\sin^2x$

we can write,

$x(1-\sin^2x)^2\:dx$

• 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,

$\displaystyle \int_{0}^{0}(1-t^2)^2\:dt$

As we know,

$\displaystyle \int_{0}^{0}f(x)\:dx=0$

So,

$\displaystyle \int_{0}^{0}(1-t^2)^2\:dt=0$

Answer :-

So, answer is Zero (0).

Additional Formulae :-

$\displaystyle \int \sin x\:dx=-\cos x+C$

$\sf{\displaystyle \int \cos x\:dx=\sin x+C}$

$\sf{\displaystyle \int \tan x\:dx=ln|\sec x|+C}$