If f ( t ) = 0 2 x ?d x 1 - cos 2 t sin 2 x , 0 < t , then the value of 0 / 2 2 ? d t f ( t ) equals ______ . [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

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]

Ans.
(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 $\tan x = v \Rightarrow \sec^{2} x d x = d v$

$When x = 0, t = 0$

$When x = \frac{π}{2}, t = \infty$

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

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

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

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

Want Us to Email you About Special Offers & Updates?