If f(t)=∫0^π (2 x d x)/(1-cos ² t sin ² x), 0<t<π, then the value of ∫0^(π/2) (π² dt)/f(t) equals __________.

Solution

Maths sir · 4 steps

Kill the x in the numerator with x → π - x, and then the tan x substitution does the rest.

1. Remove the x. Put x → π - x. Since sin²(π - x) = sin² x, the numerator becomes 2(π - x). Adding this to the original integral gives 2f(t) = ∫₀^π (2π dx)/(1-cos² t sin² x), so f(t) = π ∫₀^π dx/(1-cos² t sin² x).
2. Halve the range. The integrand depends only on sin² x, which is symmetric about x = π/2. So f(t) = 2π ∫₀^(π/2) dx/(1-cos² t sin² x).
3. Substitute tan x. Divide top and bottom by cos² x: the denominator becomes sec² x - cos² t tan² x = 1 + sin² t tan² x. With u = tan x, the integral is ∫₀^∞ du/(1 + sin² t u²) = π/(2 sin t), using sin t > 0 for 0 < t < π.

   f(t) = 2π × π/(2 sin t) = π²/(sin t)

4. So. π²/f(t) = sin t, and ∫₀^(π/2) sin t dt = [-cos t]₀^(π/2) = 1. The answer is 1.

The common slip: 2. Forgetting the factor 2 when halving ∫₀^π to ∫₀^(π/2) gives f(t) = π²/(2 sin t) and a final answer of 2.

The idea to keep. When an x sits in the numerator over a function of sin x on [0, π], use x → π - x to replace it by π/2. For 1/(a + b sin² x), divide by cos² x and put u = tan x.

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