Let [.] denote the greatest integer function.

Question

Let [.] denote the greatest integer function. Then
$\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{12(3+[x])}{3+[\sin x]+[\cos x]})dx$
is equal to :

Accepted Answer

$I=\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{12(3+[x])}{3+[\sin x]+[\cos x]})dx$

Breakdown the Interval Since $\frac{\pi }{2}\approx 1.57$ and $-\frac{\pi }{2}\approx -1.57$, we break the integral at the integer points $x=-1,0,1$.

Interval
 [x]
 [sin x]
 [cos x]
 Integrand f(x)

$[-\frac{\pi }{2},-1)$
 -2
 -1
 0

$\frac{12(3-2)}{3-1+0}=\frac{12}{2}=6$

[-1, 0)
 -1
 -1
 0

$\frac{12(3-1)}{3-1+0}=\frac{24}{2}=12$

[0, 1)
 0
 0
 0

$\frac{12(3+0)}{3+0+0}=\frac{36}{3}=12$

$[1,\frac{\pi }{2}]$
 1
 0
 0

$\frac{12(3+1)}{3+0+0}=\frac{48}{3}=16$

Notes on values :

1. For $x\in (0,1),0

To find the total value, we integrate the constant values found above over their respective subintervals:

$I=\int \_{-\pi /2}\^{-1}6dx+\int \_{-1}\^{1}12dx+\int \_{1}\^{\pi /2}16dx$

$\Rightarrow$ $I=6[x{]}\_{-1}\^{-\pi /2}+12[x{]}\_{1}\^{-1}+16[x{]}\_{\pi /2}\^{1}$

$\Rightarrow$ $I=6(-1-(-\frac{\pi }{2}))+12(1-(-1))+16(\frac{\pi }{2}-1)$

$\Rightarrow$ $

Solutions · 1