If 3 1 - sin 2 x ? d x = + 2 + 3 , where and are rational numbers, then 3 + 4 is equal to [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.
$If \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x = α + β \sqrt{2} + γ \sqrt{3}, where α, β, and γ are rational numbers, then 3 α + 4 β - γ is equal to \underline{}____.$ [2024]

Ans.
(6)

Let $I = \int_{π / 6}^{π / 3} \sqrt{1 - \sin 2 x} d x$

$= \int_{π / 6}^{π / 3} \sqrt{{(\sin x - \cos x)}^{2}} d x \Rightarrow I = \int_{π / 6}^{π / 3}$ $| \sin x - \cos x |$ $d x$

$= \int_{π / 6}^{π / 4} (\cos x - \sin x) d x + \int_{π / 4}^{π / 3} (\sin x - \cos x) d x$

$= {[\sin x + \cos x]}_{π / 6}^{π / 4} + {[- \cos x - \sin x]}_{π / 4}^{π / 3}$

$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{2} - \frac{\sqrt{3}}{2}) + (\frac{- 1}{2} - \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$

$= \frac{4}{\sqrt{2}} - 1 - \sqrt{3} = - 1 + 2 \sqrt{2} - \sqrt{3}$

$So, α = - 1, β = 2, γ = - 1$

$∴ 3 α + 4 β - γ = - 3 + 8 + 1 = 6$

Want Us to Email you About Special Offers & Updates?