JEE Main 2021 — Mathematics Calculus

We have,

$$I={\int }_{-\pi /2}^{\pi /2}([[x]-\mathrm{sin}x])dx\ldots (i)$$I=∫−π/2π/2​([[x]−sinx])dx…(i)

$$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{-1}([[x]-\mathrm{sin}x])dx+{\int }_{-1}^{0}([[x]-\mathrm{sin}x])dx+{\int }_{0}^{1}([[x]-\mathrm{sin}x])dx+{\int }_{1}^{\frac{\pi }{2}}([[x]-\mathrm{sin}x])dx$$⇒I=∫−2π​−1​([[x]−sinx])dx+∫−10​([[x]−sinx])dx+∫01​([[x]−sinx])dx+∫12π​​([[x]−sinx])dx

$$\Rightarrow I={\int }_{-\pi /2}^{-1}([-2-\mathrm{sin}x])dx+{\int }_{-1}^{0}([-1-\mathrm{sin}x])dx+{\int }_{0}^{1}([0-\mathrm{sin}x])dx+{\int }_{1}^{\frac{\pi }{2}}([1-\mathrm{sin}x])dx$$⇒I=∫−π/2−1​([−2−sinx])dx+∫−10​([−1−sinx])dx+∫01​([0−sinx])dx+∫12π​​([1−sinx])dx

$$\Rightarrow I=-2{\int }_{-\pi /2}^{-1}dx-{\int }_{-1}^{0}dx-1{\int }_{0}^{1}dx+{\int }_{1}^{\frac{\pi }{2}}0\cdot dx$$⇒I=−2∫−π/2−1​dx−∫−10​dx−1∫01​dx+∫12π​​0⋅dx

$$\Rightarrow I=-2(-1+\frac{\pi }{2})-1-1$$⇒I=−2(−1+2π​)−1−1

$$\Rightarrow I=-\pi$$⇒I=−π