If 3 1 - sin 2 x ? d x = + 2 + 3 , where and are rational numbers, then 3 + 4 is equal to [2024]

Question

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

Smart Achievers · Accepted Answer

(6)

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

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

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

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

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

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

$So, α = - 1, β = 2, γ = - 1$

$∴ 3 α + 4 β - γ = - 3 + 8 + 1 = 6$

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]

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Solution

Q. If∫π/6π/31-sin2x dx=α+β2+γ3, where α,β, and γ are rational numbers, then 3α+4β-γ is equal to ¯____. [2024]

Ans. (6) Let I=∫π/6π/31-sin2x dx =∫π/6π/3(sinx-cosx)2 dx⇒I=∫π/6π/3|sinx-cosx|dx =∫π/6π/4(cosx-sinx) dx+∫π/4π/3(sinx-cosx) dx =[sinx+cosx]π/6π/4+[-cosx-sinx]π/4π/3 =(12+12-12-32)+(-12-32+12+12) =42-1-3=-1+22-3 So, α=-1, β=2, γ=-1 ∴ 3α+4β-γ=-3+8+1=6

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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]