If ∫ cosec⁵xdx=α cot xcosecx(cosec²x+3/2)+β logₑ|tan x/2|+C where α,β∈ ℝ and C is the constant of integration, then the value of 8(α+β) equals _________.

Solution

Maths sir · 4 steps

Peel off cosec² x and integrate by parts: the power drops from 5 to 3 to 1.

1. Set up parts. Let Iₙ = ∫ cosecⁿ x dx and write the integrand as cosecⁿ⁻²x · cosec² x. Take u = cosecⁿ⁻²x and dv = cosec² x dx, so v = -cot x. Then use cot² x = cosec² x - 1 and move the Iₙ term to the left.

   Iₙ = -(cosecⁿ⁻²x cot x)/(n-1) + (n-2)/(n-1) Iₙ₋₂

2. Apply it twice. With n = 5: I₅ = -1/4cosec³ x cot x + 3/4I₃. With n = 3: I₃ = -1/2cosec x cot x + 1/2I₁, where I₁ = logₑ|tan x/2|.
3. Match the form. So I₅ = -1/4cosec³ x cot x - 3/8cosec x cot x + 3/8 logₑ|tan x/2|. Taking out -1/4 cot xcosec x gives -1/4 cot xcosec x(cosec² x + 3/2). Matching with the given form, α = -1/4 and β = 3/8.
4. So. 8(α + β) = 8(-1/4 + 3/8) = 8 × 1/8 = 1.

The common slip: 5. Losing the minus sign in front of the cot x terms makes α = 1/4, and then 8(α + β) comes out as 5, not 1.

The idea to keep. For odd powers of cosecx or sec x, peel off the square, integrate by parts and swap cot² for cosec² - 1. Each step drops the power by 2.

Answer from the NTA final answer key. Solution by PrepWiser.