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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Indefinite Integration

Let I(x)=∫ (6)/( ^2 x(1- x)^2) d x . If I(0)=3 , then I((π)/(12))...

Let I(x)=∫6sin⁡2x(1−cot⁡x)2dxI(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x. If I(0)=3I(0)=3, then $I\left(\frac{\pi}{12}\right) is equal to

Options

  1. A232 \sqrt{3}
  2. B3\sqrt{3}
  3. C333 \sqrt{3}
  4. D636 \sqrt{3}

Correct answer

C. 333 \sqrt{3}

Step-by-step solution

I(x)=∫6dxsin⁡2x(1−cot⁡x)2=∫6cosec⁡2xdx(1−cot⁡x)2I(x)=\int \frac{6 d x}{\sin ^2 x(1-\cot x)^2}=\int \frac{6 \operatorname{cosec}^2 x d x}{(1-\cot x)^2}
Put 1−cot⁡x=tcosec⁡21-\cot x=t \operatorname{cosec}^2\mathrm{xdx}=\mathrm{dt}I=∫6dtt2=−6t+cI(x)=−61−cot⁡xc,c=3I(x)=3−61−cot⁡x,I(π12)=3−61−(2+3)I(π12)=3+63+1=3+6(3−1)2=33\begin{aligned} & I=\int \frac{6 d t}{t^2}=\frac{-6}{t}+c \\ & I(x)=\frac{-6}{1-\cot x} c, c=3 \\ & I(x)=3-\frac{6}{1-\cot x}, I\left(\frac{\pi}{12}\right)=3-\frac{6}{1-(2+\sqrt{3})} \\ & I\left(\frac{\pi}{12}\right)=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3 \sqrt{3}\end{aligned}$$
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