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Question

Mathematics Question on limits and derivatives

Let I(x)=6sin2x(1cotx)2dxI(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} dx. If I(0)=3I(0) = 3, then I(π12)I\left(\frac{\pi}{12}\right) is equal to:

A

3\sqrt{3}

B

333\sqrt{3}

C

636\sqrt{3}

D

232\sqrt{3}

Answer

333\sqrt{3}

Explanation

Solution

Given : I(x)=6sin2x(1cotx)2dx.I(x) = \int \frac{6}{\sin^2 x (1 - \cot x)^2} \, dx.

Step 1: Substitution Let: t=1cotx,csc2xdx=dt.t = 1 - \cot x, \quad \csc^2 x \, dx = dt. The integral becomes: I=6dtt2=6t+c=61cotx+c.I = \int \frac{6 \, dt}{t^2} = -\frac{6}{t} + c = -\frac{6}{1 - \cot x} + c.

Step 2: Using I(0)=3I(0) = 3: At x=0x = 0, cot(0)=\cot(0) = \infty. Substituting: I(0)=3=61cot(0)+c    c=3.I(0) = 3 = -\frac{6}{1 - \cot(0)} + c \implies c = 3. Thus, the expression for I(x)I(x) becomes: I(x)=361cotx.I(x) = 3 - \frac{6}{1 - \cot x}.

Step 3 : Evaluate I(π12)I\left(\frac{\pi}{12}\right): At x=π12x = \frac{\pi}{12}: cot(π12)=2+3.\cot\left(\frac{\pi}{12}\right) = 2 + \sqrt{3}. Substitute into I(x)I(x): I(π12)=361(2+3).I\left(\frac{\pi}{12}\right) = 3 - \frac{6}{1 - (2 + \sqrt{3})}. Simplify: I(π12)=3+62+31=3+61+3.I\left(\frac{\pi}{12}\right) = 3 + \frac{6}{2 + \sqrt{3} - 1} = 3 + \frac{6}{1 + \sqrt{3}}.

Step 4 : Rationalize the denominator: 61+3=6(13)(1+3)(13)=6(13)13=6(31)2=3(31).\frac{6}{1 + \sqrt{3}} = \frac{6(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{6(1 - \sqrt{3})}{1 - 3} = \frac{6(\sqrt{3} - 1)}{2} = 3(\sqrt{3} - 1). Substitute back: I(π12)=3+333=33.I\left(\frac{\pi}{12}\right) = 3 + 3\sqrt{3} - 3 = 3\sqrt{3}.

Final Answer: 33.\boxed{3\sqrt{3}.}