int0 0- pi - 2 sqrt 1-sin 2 x - dxa 2 sqrt 2 b 2 sqrt 2-1c 2d 2 sqrt 2-1

I = ∫_0^π/2√(1 sin2x) dx = ∫_0^π/4√((cosx sinx)^2) dx+∫_π/4^π/2√((sinx cosx)^2)dx =[sinx + cosx]_0^π/4 + [ cosx sinx]_π/4^π/2 =1/√(2)+1/√(2) 0 1+( 0 1 ...

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Byju's Answer

Standard XII

Mathematics

Inequality

int0 0∧ pi / ...

Question

$\int_{0}^{π / 2} \sqrt{1 - s i n 2 x} d x (a) 2 \sqrt{2} (b) 2 (\sqrt{2} + 1) (c) 2 (d) 2 (\sqrt{2} - 1)$

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Solution

$I = \int_{0}^{π / 2} \sqrt{1 - s i n 2 x} d x = \int_{0}^{π / 4} \sqrt{(c o s x - s i n x)^{2}} d x + \int_{π / 4}^{π / 2} \sqrt{(s i n x - c o s x)^{2}} d x = [s i n x + c o s x]_{0}^{π / 4} + [- c o s x - s i n x]_{\frac{π}{4}}^{\frac{π}{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 0 - 1 + (- 0 - 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 2 \sqrt{2} - 2 = 2 (\sqrt{2 - 1})$

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[Karnataka CET 1999]

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∫π/20√1−sin2x dx(a)2√2(b)2(√2+1)(c)2(d)2(√2−1)

I=∫π/20√1−sin2x dx=∫π/40√(cosx−sinx)2dx+∫π/2π/4√(sinx−cosx)2dx=[sinx+cosx]π/40+[−cosx−sinx]π2π4=1√2+1√2−0−1+(−0−1+1√2+1√2)=2√2−2=2(√2−1)

$\int_{0}^{π / 2} \sqrt{1 - s i n 2 x} d x (a) 2 \sqrt{2} (b) 2 (\sqrt{2} + 1) (c) 2 (d) 2 (\sqrt{2} - 1)$