JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
The value of ∫π/3^(π/2) (2+3 sin x)/(sin x(1+cos x)) d x is equal to :
- (a) 10/3-√3+log ₑ √3
- (b) 7/2-√3-log ₑ √3
- (c) 10/3-√3-log ₑ √3
- (d) -2+3√3+log ₑ √3
Answer and solution
Answer (a) 10/3-√3+log ₑ √3
Solution
Maths sir · 4 steps
Put t = tan x/2 and the whole messy integrand turns into 1/t + t + 3.
- Half-angle substitution. Let t = tan x/2. Then sin x = 2t/(1+t²), 1+cos x = 2/(1+t²) and dx = (2 dt)/(1+t²). The limits change from x = π/3 to t = 1/√3, and from x = π/2 to t = 1.
- Split and simplify. Write the integrand as 2/(sin x(1+cos x)) + 3/(1+cos x). After the substitution, the first part times dx becomes (1+t²)/t dt = (1/t + t)dt, and the second becomes 3 dt.
I = ∫_(1/√3)¹ (1/t + t + 3)dt = [ln t + t²/2 + 3t]_(1/√3)¹
- Put in limits. At t = 1 the bracket is 0 + 1/2 + 3 = 7/2. At t = 1/√3 it is -ln√3 + 1/6 + √3. Subtracting gives 7/2 - 1/6 - √3 + ln√3.
- So. I = 10/3 - √3 + logₑ√3, which is option (a).
The idea to keep. When an integrand is built only from sin x and cos x, try t = tan x/2. Factors like 1+cos x collapse neatly, and you are left with a simple algebraic integral.
Answer from the NTA final answer key. Solution by PrepWiser.