Let I(x)=∫ 6/(sin² x(1-cot x)²)dx. If I(0)=3, then I(π/12) is equal to

Solution

Maths sir · 5 steps

Pull out csc² x: it is exactly the derivative of 1-cot x, so substitute that.

1. Spot the substitution. Since 1/(sin² x) = csc² x, the integrand is (6 csc² x)/(1-cot x)². Put t = 1-cot x. Then dt = csc² x dx.

   I = ∫ 6/t² dt = -6/t + C = -6/(1-cot x) + C

2. Rewrite in tan. cot 0 is undefined, so first rewrite the answer in terms of tan x. Multiply the top and bottom by sin x, then divide both by cos x: -(6 sin x)/(sin x - cos x) = (6 tan x)/(1-tan x). So I(x) = (6 tan x)/(1-tan x) + C.
3. Find C. At x = 0, tan 0 = 0, so I(0) = C = 3.
4. Evaluate at the point. tan π/12 = 2-√3 and 1-tan π/12 = √3-1. Rationalising gives (2-√3)/(√3-1) = (2-√3)(√3+1)/2 = (√3-1)/2. So the first term is 6 × (√3-1)/2 = 3√3-3.
5. So. I(π/12) = 3√3-3+3 = 3√3, which is option (d).

The idea to keep. When csc² x multiplies a function of cot x, put t = cot x or t = 1-cot x. If the given condition is at a point where cot blows up, switch to tan first.

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