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JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Indefinite Integrals

Let

I (x) = \int \frac{6}{s i n ^{2} x ( 1 - c o t x ) ^{2}} dx

. If

I (0) = 3

, then

I (\frac{π}{12})

is equal to :

(A) $3$
(B) $33$ ✓ Correct
(C) $63$
(D) $23$

Answer: (B)

Solution

Put

t = 1 - cot x

,

d t = csc^{2} x d x

; then

I = \int \frac{6 d t}{t ^{2}} = - \frac{6}{1 - cot x} + C

. The condition fixes

C

(as

x \to 0^{+}

,

cot x \to \infty

so the first term

\to 0

, giving

C = 3

). With

cot \frac{π}{12} = 2 + 3

:

I (\frac{π}{12}) = 3 - \frac{6}{1 - ( 2 + 3 )} = 3 + \frac{6}{1 + 3} = 3 + 3 (3 - 1) = 33

. Option (2).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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