Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2009 · Shift 0 · Q33
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2009 · Shift 0 · Q33

Definite Integration question

2009 · Shift 0 · Q33

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
∫0π[cot⁡x]dx,\int\limits_0^\pi {\left[ {\cot x} \right]dx,}0∫π​[cotx]dx, where [.]\left[ . \right][.] denotes the greatest integer function, is equal to:
  1. A
    111
  2. B
    −1-1−1
  3. C
    −π2- {\pi \over 2}−2π​
  4. D
    π2{\pi \over 2}2π​
View written solutionFree

Correct answer: C

  1. Understand the integrand

We need to evaluate

I=∫0π[cot⁡x] dx,I=\int_0^\pi [\cot x]\,dx,I=∫0π​[cotx]dx,

where [t][t][t] denotes the greatest integer less than or equal to ttt.

  1. Use symmetry of cotangent

Recall:

cot⁡(π−x)=−cot⁡x.\cot(\pi-x)=-\cot x.cot(π−x)=−cotx.

Hence,

[cot⁡(π−x)]=[−cot⁡x].[\cot(\pi-x)]=[-\cot x].[cot(π−x)]=[−cotx].

So let us pair the integrand at xxx and π−x\pi-xπ−x.

Then

I=∫0π/2[cot⁡x]dx+∫π/2π[cot⁡x]dx.I=\int_0^{\pi/2}[\cot x]dx+\int_{\pi/2}^{\pi}[\cot x]dx.I=∫0π/2​[cotx]dx+∫π/2π​[cotx]dx.

In the second integral, put x=π−tx=\pi-tx=π−t:

∫π/2π[cot⁡x]dx=∫π/20[cot⁡(π−t)](−dt)=∫0π/2[−cot⁡t]dt.\int_{\pi/2}^{\pi}[\cot x]dx= \int_{\pi/2}^{0}[\cot(\pi-t)](-dt) =\int_0^{\pi/2}[-\cot t]dt.∫π/2π​[cotx]dx=∫π/20​[cot(π−t)](−dt)=∫0π/2​[−cott]dt.

Therefore,

I=∫0π/2([cot⁡x]+[−cot⁡x]) dx.I=\int_0^{\pi/2}\big([\cot x]+[-\cot x]\big)\,dx.I=∫0π/2​([cotx]+[−cotx])dx.
  1. Use the floor-function identity

For any real uuu which is not an integer,

[u]+[−u]=−1.[u]+[-u]=-1.[u]+[−u]=−1.

Now for x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2), cot⁡x>0\cot x>0cotx>0 and varies continuously from +∞+\infty+∞ to 000. It becomes an integer only at isolated points, and such points do not affect the value of the integral.

Thus, for almost every x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2),

[cot⁡x]+[−cot⁡x]=−1.[\cot x]+[-\cot x]=-1.[cotx]+[−cotx]=−1.

Hence,

I=∫0π/2(−1) dx=−π2.I=\int_0^{\pi/2}(-1)\,dx=-\frac{\pi}{2}.I=∫0π/2​(−1)dx=−2π​.
  1. Check endpoints / improper behavior

Although cot⁡x→+∞\cot x\to +\inftycotx→+∞ as x→0+x\to 0^+x→0+ and cot⁡x→−∞\cot x\to -\inftycotx→−∞ as x→π−x\to \pi^-x→π−, the paired symmetry above gives a well-defined improper integral value, and isolated points where cot⁡x∈Z\cot x\in\mathbb Zcotx∈Z do not change the integral.

  1. Evaluate options
  • A: 111 — incorrect
  • B: −1-1−1 — incorrect
  • C: −π2-\dfrac{\pi}{2}−2π​ — correct
  • D: π2\dfrac{\pi}{2}2π​ — incorrect

So the value is

−π2.\boxed{-\frac{\pi}{2}}.−2π​​.
PreviousNext

More from Definite Integration

  • Let I=0∫1​x​sinx​dx and J=0∫1​x​cosx​dx. Then which one of the following is true?2007 · MCQ
  • The solution for x of the equation 2​∫x​tt2−1​dt​=2π​ is2007 · MCQ
  • Let F(x)=f(x)+f(x1​), where f(x)=l∫x​1+tlogt​dt, Then F(e) equals2007 · MCQ
  • The value of 1∫a​[x]f′(x)dx,a>1 where [x] denotes the greatest integer not exceeding x is2006 · MCQ
  • 0∫π​xf(sinx)dx is equal to2006 · MCQ
  • −23π​∫−2π​​[(x+π)3+cos2(x+3π)]dx is equal to2006 · MCQ
  • Let f:R→R be a differentiable function having f(2)=6, f′(2)=(481​). Then x→2lim​6∫f(x)​x−24t3​dt…2005 · MCQ
  • The value of −π∫π​1+axcos2​dx,a>0, is2005 · MCQ