If - sin 2x -cos 2x dx- 1 - 2 sin 2x-a -b- then

The correct option is A a=π/ 4 ,b=0 ∫ (sin2x cos2x)dx = 1√(2)sin (2x a)+b cos2x2 sin2x2 = 1√(2)sin(2x a)+b = 1√(2) sin2xcosa 1√(2)sina cos ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

If ∫ sin 2x...

Question

If $\int (sin 2 x - cos 2 x) d x = \frac{1}{\sqrt{2}} sin (2 x - a) + b,$ , then

a = \frac{π}{4}, b = 0

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a = - \frac{π}{4}, b = 0

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a = \frac{5 π}{4}, b =

any constant

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a = - \frac{5 π}{4}, b =

any constant

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Solution

The correct option is A $a = \frac{π}{4}, b = 0$
$\int (s i n 2 x - c o s 2 x) d x = \frac{1}{\sqrt{2}} s i n (2 x - a) + b$
$\frac{- c o s 2 x}{2} - \frac{s i n 2 x}{2} = \frac{1}{\sqrt{2}} s i n (2 x - a) + b$
$= \frac{1}{\sqrt{2}} s i n 2 x c o s a - \frac{1}{\sqrt{2}} s i n a c o s 2 x + b$
$\frac{1}{\sqrt{2}} s i n 2 x c o s a = \frac{- s i n 2 x}{2}$
$c o s a = \frac{- 1}{\sqrt{2}}$
$a = \frac{π}{4} .$
$\frac{- 1}{\sqrt{2}} s i n a c o s 2 x = \frac{- c o s 2 x}{2}$
$s i n a = \frac{1}{\sqrt{2}}$
$a = \frac{π}{y}$
$b = 0$
$∴ a = \frac{π}{4}, b = 0$

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If ∫(sin2x−cos2x)dx=1√2sin(2x−a)+b,, then

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