If f ( t ) = 0 2 x ?d x 1 - cos 2 t sin 2 x , 0

Question

If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]

Smart Achievers · Accepted Answer

(1)

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

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

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

Adding (i) and (ii), we get

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

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

On dividing the numerator and denominator by $\cos^{2} x,$ we get

$f (t) = 2 π \int_{0}^{π / 2} \frac{\sec^{2} x d x}{\sec^{2} x - \cos^{2} t \tan^{2} x}$

Put $\tan x = v \Rightarrow \sec^{2} x d x = d v$

$When x = 0, t = 0$

$When x = \frac{π}{2}, t = \infty$

$∴ f (t) = 2 π \int_{0}^{\infty} \frac{d v}{v^{2} + 1 - \cos^{2} t v^{2}}$

$= 2 π \int_{0}^{\infty} \frac{d v}{1 + {(v \sin t)}^{2}} = 2 π {[\frac{\tan^{- 1} (v \sin t)}{\sin t}]}_{0}^{\infty}$

$= 2 π \frac{π}{2 \sin t} = \frac{π^{2}}{\sin t}$

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

Solution

Q.
If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]

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Solution

Q. If f(t)=∫0π2x dx1-cos2tsin2x,0<t<π, then the value of ∫0π/2π2 dtf(t) equals ______ . [2024]

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Q.
If $f (t) = \int_{0}^{π} \frac{2 x d x}{1 - \cos^{2} t \sin^{2} x}, 0 < t < π,$ then the value of $\int_{0}^{π / 2} \frac{π^{2} d t}{f (t)}$ equals ______ . [2024]