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

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

Question

[\,\cdot\,]

denotes the greatest integer function,

\displaystyle\int_{0}^{2\pi}\big[\,|\sin x|-|\cos x|\,\big]\,dx

equals

-2\pi

-4\pi

-\pi

correct

0

Solution

Step 1: Note that

|\sin x|-|\cos x|

is periodic with period

\pi

and takes values in

[-1,1]

. Over

[0,2\pi]

there are two such periods, so

\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,\pi]

the quantity

h(x)=|\sin x|-|\cos x|

is negative near the ends and reaches

1

only at the single point

x=\dfrac{\pi}{2}

. Precisely,

h(x)\in(-1,0)

\left(0,\dfrac{\pi}{4}\right)\cup\left(\dfrac{3\pi}{4},\pi\right)

, while

h(x)\in[0,1)

\left(\dfrac{\pi}{4},\dfrac{3\pi}{4}\right)

. Thus

[h(x)]=-1

on the outer parts and

[h(x)]=0

on the middle part. Step 3: Evaluate the integral over one period:

\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:

\int_{0}^{2\pi}[h(x)]\,dx=2\left(-\dfrac{\pi}{2}\right)=-\pi.

Correct answer: (3)

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