If ∫ (sin x)/(sin³x+cos³x)dx=
αlogₑ|1+tan x|+βlogₑ|1-tan x+tan²x|+γtan⁻¹((2 tan x-1)/√3)+C, when C is constant of integration, then the value of 18(α+β+γ²) is ______________.

Solution

Maths sir · 4 steps

Divide top and bottom by cos³ x, put t = tan x, and you have a plain rational function in t.

1. Change to tan x. Divide numerator and denominator by cos³ x. The integrand becomes (tan x sec² x)/(tan³ x + 1), and with t = tan x, dt = sec² x dx, the integral is ∫ t/(t³+1) dt.

   t/(t+1)(t²-t+1) = (-1/3)/(t+1) + (1/3(t+1))/(t²-t+1)

2. Find the constants. Put t = -1 in t = A(t²-t+1) + (Bt+C)(t+1) to get A = -1/3. Matching the t² terms gives B = -A = 1/3, and matching constants gives C = -A = 1/3. The first part integrates to -1/3 ln|1+tan x|, so α = -1/3.
3. Split the numerator. Write t+1 = 1/2(2t-1) + 3/2. The first piece gives 1/3 × 1/2 ln|t²-t+1|, so β = 1/6. The second gives 1/3 × 3/2 × 2/√3 tan⁻¹(2t-1)/√3, so γ = 1/√3.
4. So. α + β + γ² = -1/3 + 1/6 + 1/3 = 1/6, and 18 × 1/6 = 3.

The common slip: 15. Taking α as +1/3 instead of -1/3 makes the sum 5/6 and gives 15, not 3. At t = -1 the numerator t is -1, so A is negative.

The idea to keep. When the denominator has degree 3 in sin x and cos x and the numerator has degree 1, divide by cos³ x and set t = tan x. Then use partial fractions in t.

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