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MTH 102 MTH 102 Past Questions Objective Question
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Question

Which of the following option best describes the reduction formula for Vn = ∫ cosecn x dx, n ≥ 2.
Options
A -<sup>1</sup>/<sub>(n-1)</sub> cosec<sup>n-2</sup>x cot x + <sup>(n-2)</sup>/<sub>(n-1)</sub> V<sub>n-2</sub>
Correct Answer
B -<sup>1</sup>/<sub>(1-n)</sub> cosec<sup>n-2</sup>x cot x - <sup>(n-2)</sup>/<sub>(n-1)</sub> V<sub>n-2</sub>
C -<sup>1</sup>/<sub>(n-1)</sub> cosec<sup>n-2</sup>x tan x + <sup>(n-2)</sup>/<sub>(n-1)</sub> V<sub>n-2</sub>
D None.
Correct Answer

Option A is the correct answer.

Detailed Explanation

The reduction formula for ∫ cosecnx dx is derived using integration by parts. Let u = cosecn-2x and dv = cosec2x dx. Then du = (n-2)cosecn-3x (-cosec x cot x) dx = -(n-2)cosecn-2x cot x dx, and v = -cot x.Applying ∫u dv = uv - ∫v du:∫cosecnx dx = cosecn-2x (-cot x) - ∫(-cot x)(-(n-2)cosecn-2x cot x) dx= -cosecn-2x cot x - (n-2)∫cosecn-2x cot²x dxUse the identity cot²x = cosec²x - 1:= -cosecn-2x cot x - (n-2)∫cosecn-2x (cosec²x - 1) dx= -cosecn-2x cot x - (n-2)∫cosecnx dx + (n-2)∫cosecn-2x dxLet In = ∫cosecnx dx.In = -cosecn-2x cot x - (n-2)In + (n-2)In-2(1 + n - 2)In = -cosecn-2x cot x + (n-2)In-2(n-1)In = -cosecn-2x cot x + (n-2)In-2In = -1/(n-1) cosecn-2x cot x + (n-2)/(n-1) In-2. This matches option A.

Hint

Use integration by parts with u = cosecn-2x and dv = cosec2x dx, then use the identity cot²x = cosec²x - 1.

Question Info

Course: MTH 102
Topic: MTH 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 261 of 930
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