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Indefinite Integrals question

2017 · 8 Apr · Shift 1 · Q48
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Indefinite Integrals question

2017 · 8 Apr · Shift 1 · Q48

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
The integral ∫1+2cot⁡x(csc⁡x+cot⁡x) dx(0<x<π2)\int \sqrt{1 + 2\cot x(\csc x + \cot x)} \, dx \quad \left(0 \lt x \lt \frac{\pi}{2}\right)∫1+2cotx(cscx+cotx)​dx(0<x<2π​) is equal to : (where C is a constant of integration)
  1. A
    4 log(sin x2{x \over 2}2x​ ) + C
  2. B
    2 log(sin x2{x \over 2}2x​ ) + C
  3. C
    2 log(cos x2{x \over 2}2x​ ) + C
  4. D
    4 log(cos x2{x \over 2}2x​) + C
View written solutionFree

Correct answer: B

  1. Simplify the expression inside the square root

We need to evaluate

I=∫1+2cot⁡x(csc⁡x+cot⁡x) dx,0<x<π2.I=\int \sqrt{1+2\cot x(\csc x+\cot x)}\,dx, \qquad 0<x<\frac\pi2.I=∫1+2cotx(cscx+cotx)​dx,0<x<2π​.

Expand the term inside:

1+2cot⁡x(csc⁡x+cot⁡x)=1+2cot⁡xcsc⁡x+2cot⁡2x.1+2\cot x(\csc x+\cot x)=1+2\cot x\csc x+2\cot^2 x.1+2cotx(cscx+cotx)=1+2cotxcscx+2cot2x.

Now use the identity

csc⁡2x=1+cot⁡2x.\csc^2 x=1+\cot^2 x.csc2x=1+cot2x.

Observe that

n=(1+cot⁡2x)+2cot⁡xcsc⁡x+cot⁡2x=csc⁡2x+2cot⁡xcsc⁡x+cot⁡2x.n= (1+\cot^2 x)+2\cot x\csc x+\cot^2 x = \csc^2 x+2\cot x\csc x+\cot^2 x.n=(1+cot2x)+2cotxcscx+cot2x=csc2x+2cotxcscx+cot2x.

This is a perfect square:

csc⁡2x+2cot⁡xcsc⁡x+cot⁡2x=(csc⁡x+cot⁡x)2.\csc^2 x+2\cot x\csc x+\cot^2 x=(\csc x+\cot x)^2.csc2x+2cotxcscx+cot2x=(cscx+cotx)2.

Hence,

1+2cot⁡x(csc⁡x+cot⁡x)=∣csc⁡x+cot⁡x∣.\sqrt{1+2\cot x(\csc x+\cot x)}=|\csc x+\cot x|.1+2cotx(cscx+cotx)​=∣cscx+cotx∣.

  1. Remove modulus using the given interval

Given 0<x<π20<x<\frac\pi20<x<2π​, we have sin⁡x>0\sin x>0sinx>0 and cos⁡x>0\cos x>0cosx>0, so

csc⁡x>0,cot⁡x>0.\csc x>0,\qquad \cot x>0.cscx>0,cotx>0.

Therefore,

∣csc⁡x+cot⁡x∣=csc⁡x+cot⁡x.|\csc x+\cot x|=\csc x+\cot x.∣cscx+cotx∣=cscx+cotx.

So the integral becomes

I=∫(csc⁡x+cot⁡x) dx.I=\int (\csc x+\cot x)\,dx.I=∫(cscx+cotx)dx.

  1. Use a standard identity

Recall the half-angle identity:

csc⁡x+cot⁡x=1+cos⁡xsin⁡x.\csc x+\cot x=\frac{1+\cos x}{\sin x}.cscx+cotx=sinx1+cosx​.

Now,

=\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} =\cot\frac x2.$$ Thus, $$I=\int \cot\frac x2\,dx.$$ 4. **Integrate** Let $$u=\frac x2 \Rightarrow dx=2\,du.$$ Then $$I=2\int \cot u\,du=2\log(\sin u)+C.$$ Substituting back $u=\frac x2$: $$I=2\log\left(\sin\frac x2\right)+C.$$ 5. **Match with the options** This is exactly: $$\boxed{2\log\left(\sin\frac x2\right)+C}$$ So the correct option is **B**.
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