The value of - 2 y ( 1 + sin y ) 1 + cos 2 y ? d y is: [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.
The value of $\int_{- π}^{π} \frac{2 y (1 + \sin y)}{1 + \cos^{2} y} d y$ is: [2024]

1 $π^{2}$

2 $\frac{π^{2}}{2}$

3 $\frac{π}{2}$

4 $2 π^{2}$

Ans.
(1)

$Let I = \int_{- π}^{π} \frac{2 y (1 + \sin y)}{1 + \cos^{2} y} d y$

$I = \underset{Odd function}{\underset{⏟}{\int_{- π}^{π} \frac{2 y}{1 + \cos^{2} y} d y}} + \underset{Even function}{\underset{⏟}{\int_{- π}^{π} \frac{2 y \sin y}{1 + \cos^{2} y} d}}$

$∴$ $I = 2 \int_{0}^{π} \frac{2 y \sin y}{1 + \cos^{2} y} d y$ ...(i)

$Now, I = 4 \int_{0}^{π} \frac{(π - y) \sin (π - y)}{1 + \cos^{2} (π - y)} d y$

$\Rightarrow I = 4 \int_{0}^{π} \frac{(π - y) \sin y}{1 + \cos^{2} y} d y$ ...(ii)

$Adding (i) and (ii), we get$

$2 I = \int_{0}^{π} \frac{4 π \sin y}{1 + \cos^{2} y} d y$

$Let \cos y = t \Rightarrow \sin y d y = - d t$

$When y = 0, t = 1$

$when y = π, t = - 1$

$∴ I = - 2 π \int_{1}^{- 1} \frac{1}{1 + t^{2}} d t$

$= - 2 π {[\tan^{- 1} t]}_{1}^{- 1} = - 2 π [\frac{- π}{4} - \frac{π}{4}]$

$= - 2 π (\frac{- π}{2}) = π^{2}$

Want Us to Email you About Special Offers & Updates?