pi-pi-cos-3-x-dx-

Question Number 95549 by Fikret last updated on 25/May/20

\int_{- π}^{π} ∣ {c o s}^{3} x ∣ d x

Answered by MJS last updated on 26/May/20

\underset{- π}{\int^{π}} ∣ \cos^{3} x ∣ d x = 4 \underset{0}{\int^{π / 2}} \cos^{3} x d x =

= \underset{0}{\int^{π / 2}} (\cos 3 x + 3 \cos x) d x =

= {[\frac{1}{3} \sin 3 x + 3 \sin x]}_{0}^{π / 2} = \frac{8}{3}

Commented by peter frank last updated on 26/May/20

why you change limit sir

Commented by MJS last updated on 26/May/20

because ∣ \cos^{3} x ∣ is periodic \dots scetch it

Commented by peter frank last updated on 26/May/20

how abou this

\int_{- π}^{π} ∣ \sin x ∣ dx ?

Commented by MJS last updated on 26/May/20

\underset{- π}{\int^{π}} ∣ \sin x ∣ d x = 4 \underset{0}{\int^{π / 2}} \sin x d x

the idea is to get rid of the absolute by

finding an interval where the function is

greater than or equal to zero

Commented by peter frank last updated on 26/May/20

thank you

Answered by 1549442205 last updated on 26/May/20

I = \int_{- Π}^{Π} ∣ \cos^{3} x ∣ dx = \int_{- Π}^{\frac{- Π}{2}} (- \cos^{3} x) dx

+ \int_{- \frac{Π}{2}}^{\frac{Π}{2}} \cos^{3} xdx + \int_{\frac{Π}{2}}^{Π} (- \cos^{3} x) dx

We have F (x) = \int \cos^{3} xdx = \int (1 - \sin^{2} x) dsinx

= sinx - \frac{\sin^{3} x}{3} + C . Therefore,

I = F (- Π) - F (\frac{- Π}{2}) + F (\frac{Π}{2}) - F (\frac{- Π}{2})

+ F (\frac{Π}{2}) - F (Π) = 2 F (\frac{Π}{2}) - 2 F (\frac{- Π}{2})

= \frac{4}{3} - 2 (\frac{- 2}{3}) = \frac{8}{3} (Since F (Π) = F (- Π) = 0

. Thus, I = \frac{8}{3}

Answered by Ar Brandon last updated on 26/May/20

\int_{- π}^{π} ∣ \cos^{3} (x) ∣ dx = 4 xI; where I = \int_{0}^{\frac{π}{2}} \cos^{3} (x) dx = \frac{2}{3} {{Walli}^{'} s method}

\Rightarrow \int_{- π}^{π} ∣ \cos^{3} (x) ∣ dx = 4 \times \frac{2}{3} = \frac{8}{3}