Visual SolutionPYQ 2024 · Apr Shift 1Tricky
Question
Let I(x)=∫6sin⁡2x(1−cot⁡x)2 dxI(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)232\sqrt{3}
(B)3\sqrt{3}
(C)333\sqrt{3}
(D)636\sqrt{3}
Solution Path
Sub t=1−cot⁡xt = 1-\cot x gives I=−6/(1−cot⁡x)+3I = -6/(1-\cot x) + 3. At x=π/12x = \pi/12: cot⁡=2+3\cot = 2+\sqrt{3}, so I=33I = 3\sqrt{3}.
01Question Setup
1/4
I(x)=∫6sin⁡2x (1−cot⁡x)2 dxI(x) = \int \frac{6}{\sin^2 x\,(1-\cot x)^2}\,dx with I(0)=3I(0) = 3. Find I(π12)I(\frac{\pi}{12}).
I(π12)=  ?I(\frac{\pi}{12}) = \;?
02Substitution
2/4
Let t=1−cot⁡xt = 1 - \cot x, dt=csc⁡2x dxdt = \csc^2 x\,dx. Integral becomes −61−cot⁡x+C-\frac{6}{1-\cot x} + C.
I=−61−cot⁡x+CI = \frac{-6}{1-\cot x} + C
03Apply ConditionsKEY INSIGHT
3/4
I(0)=3⇒C=3I(0) = 3 \Rightarrow C = 3. With cot⁡π12=2+3\cot\frac{\pi}{12} = 2+\sqrt{3}: I=61+3+3=33I = \frac{6}{1+\sqrt{3}} + 3 = 3\sqrt{3}.
333\sqrt{3}
04Final Answer
4/4
I(π12)=33I(\frac{\pi}{12}) = 3\sqrt{3}.
33\boxed{3\sqrt{3}}
Concepts from this question1 concepts unlocked
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Substitution Method

EASY

Replace a composite expression with a single variable to simplify the integrand

∫f(g(x)) g′(x) dx=∫f(t) dt,t=g(x)\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad t = g(x)

Transforms complex integrands into standard forms that can be integrated directly

Trigonometric integralsRational functionsPower rule integrals
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