If f(t)=∫_0^π 2 x ~d x 1- ^2 t ^2 x , 0 < t < π , then the value of ∫_0^ (π)/(2) π^2 dt f(t) equals_________

Correct answer: 0 f(t)=∫_0^π (2 x)/(1- ^2 t ^2 x) d x .....(1) =2 ∫_0^π ((π-x) d x)/(1- ^2 ^2 x) .....(2) aligned & 2 f(t)=2 ∫_0^π (π)/(1- ^2 ^2 x) d x & f(t)=∫_0^π (π)/(1- ^2 t ^2 x) d x aligned divide \& by ^2 x aligned & f(t)=π ∫_0^π ( ^2 x d x)/( ^2 x- ^2 t^2 x) & f(t)=2 π ∫_0^ π / 2 ( ^2 x d x)/( ^2 x- ^2 t^2 ^2 x) & x=z & ^2 x d x=d z & f(t)=2 π ∫_0^ ∞ (d z)/(1+ ^2 t · z^2) & =(π^2)/( t) aligned Then ∫_0^ π / 2 (π^2)/(f(t)) d t aligned & =∫_0^ π / 2 t d t & =1 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Definite Integration

If f(t)=∫_0^π 2 x ~d x 1- ^2 t ^2 x , 0 < t < π , then the value of...

f(t)=\int_0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0 < \mathrm{t} < \pi

, then the value of

\int_0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}

equals_________

Correct answer

Step-by-step solution

f(t)=\int_0^\pi \frac{2 x}{1-\cos ^2 t \sin ^2 x} d x

.....(1)

=2 \int_0^\pi \frac{(\pi-x) d x}{1-\cos ^2 \sin ^2 x}

.....(2)

\begin{aligned} & 2 f(t)=2 \int_0^\pi \frac{\pi}{1-\cos ^2 \sin ^2 x} d x \\ & f(t)=\int_0^\pi \frac{\pi}{1-\cos ^2 t \sin ^2 x} d x \end{aligned}

divide \& by

\cos ^2 \mathrm{x}

\begin{aligned} & f(t)=\pi \int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 x} \\ & f(t)=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 \tan ^2 x} \\ & \tan x=z \\ & \sec ^2 x d x=d z \\ & f(t)=2 \pi \int_0^{\infty} \frac{d z}{1+\sin ^2 t \cdot z^2} \\ & =\frac{\pi^2}{\sin t} \end{aligned}

Then $\int_0^{\pi / 2} \frac{\pi^2}{f(t)} d t

\begin{aligned} & =\int_0^{\pi / 2} \sin t d t \\ & =1 \end{aligned}$$$

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