If \(f(t)=\int_{0}^{\pi} \frac{2x\,dx}{1-\cos^{2}t \sin^{2}x}, 0<t<\pi\), then the value of \(\int_{0}^{\frac{\pi}{2}} \frac{\pi^{2}\,dt}{f(t)}\) equals ____ .

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Definite Integrals

f (t) = \int_{0}^{π} \frac{2 x d x}{1 - c o s ^{2} t s i n ^{2} x}, 0 < t < π

, then the value of

\int_{0}^{\frac{π}{2}} \frac{π ^{2} d t}{f ( t )}

equals ____ .

Answer: 1

f (t) = \int_{0}^{π} \frac{2 x d x}{1 - cos ^{2} t sin ^{2} x}

Using the substitution

x \to π - x

(King's rule) on the same integral,

f (t) = \int_{0}^{π} \frac{2 ( π - x ) d x}{1 - cos ^{2} t sin ^{2} x}

Adding the two forms of

f (t)

2 f (t) = 2 π \int_{0}^{π} \frac{d x}{1 - cos ^{2} t sin ^{2} x} = 4 π \int_{0}^{π /2} \frac{d x}{1 - cos ^{2} t sin ^{2} x}

(using the symmetry of the integrand about

x = π /2

). Dividing numerator and denominator by

cos^{2} x

and substituting

z = tan x

f (t) = 2 π \int_{0}^{π /2} \frac{sec ^{2} x d x}{sec ^{2} x - cos ^{2} t tan ^{2} x} = 2 π \int_{0}^{\infty} \frac{d z}{1 + sin ^{2} t z ^{2}} = 2 π \cdot \frac{π /2}{sin t} = \frac{π ^{2}}{sin t}

\int_{0}^{π /2} \frac{π ^{2}}{f ( t )} d t = \int_{0}^{π /2} sin t d t = [- cos t]_{0}^{π /2} = 1

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Integer answer