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GIF Integral ∫[|sinx|−|cosx|] dx, 0 to 2π | JEE

JEE Maths question with a full step-by-step solution.

Question
If [ ⋅ ][\,\cdot\,] denotes the greatest integer function, ∫02π[ ∣sin⁡x∣−∣cos⁡x∣ ] dx\displaystyle\int_{0}^{2\pi}\big[\,|\sin x|-|\cos x|\,\big]\,dx equals
A−2π-2\pi
B−4π-4\pi
C−π-\picorrect
D00
Solution
Step 1: Note that ∣sin⁡x∣−∣cos⁡x∣|\sin x|-|\cos x| is periodic with period π\pi and takes values in [−1,1][-1,1]. Over [0,2π][0,2\pi] there are two such periods, so
∫02π[ ∣sin⁡x∣−∣cos⁡x∣ ] dx=2∫0π[ ∣sin⁡x∣−∣cos⁡x∣ ] dx.\int_{0}^{2\pi}\big[\,|\sin x|-|\cos x|\,\big]\,dx=2\int_{0}^{\pi}\big[\,|\sin x|-|\cos x|\,\big]\,dx.
 Step 2: On [0,π][0,\pi] the quantity h(x)=∣sin⁡x∣−∣cos⁡x∣h(x)=|\sin x|-|\cos x| is negative near the ends and reaches 11 only at the single point x=π2x=\dfrac{\pi}{2}. Precisely, h(x)∈(−1,0)h(x)\in(-1,0) on (0,π4)∪(3π4,π)\left(0,\dfrac{\pi}{4}\right)\cup\left(\dfrac{3\pi}{4},\pi\right), while h(x)∈[0,1)h(x)\in[0,1) on (π4,3π4)\left(\dfrac{\pi}{4},\dfrac{3\pi}{4}\right). Thus [h(x)]=−1[h(x)]=-1 on the outer parts and [h(x)]=0[h(x)]=0 on the middle part. Step 3: Evaluate the integral over one period:
∫0π[h(x)] dx=∫0π/4(−1) dx+∫π/43π/40 dx+∫3π/4π(−1) dx=−π4+0−π4=−π2.\int_{0}^{\pi}[h(x)]\,dx=\int_{0}^{\pi/4}(-1)\,dx+\int_{\pi/4}^{3\pi/4}0\,dx+\int_{3\pi/4}^{\pi}(-1)\,dx=-\dfrac{\pi}{4}+0-\dfrac{\pi}{4}=-\dfrac{\pi}{2}.
 Step 4: Double it:
∫02π[h(x)] dx=2(−π2)=−π.\int_{0}^{2\pi}[h(x)]\,dx=2\left(-\dfrac{\pi}{2}\right)=-\pi.
 Correct answer: (3)
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