- -sin-2x-cos-2x-dx - 1-2-sin-2x-a-c then a -5 -4-5 -4-43 -4

The correct option is B 5 π/4∫ (sin 2x cos 2x) dx = 1/2[cos 2x+sin 2x]+c = 1/2√(2). sin ( 2x+π/4 )+c = 1/√(2)sin ( 2x+π/4 ) =1/√(2) sin ( 2x+5 π/4 ) ⇒α ...

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Standard XII

Mathematics

Integration as Antiderivative

∫ (sin 2x-cos...

Question

$\int (s i n 2 x - c o s 2 x) d x = \frac{1}{\sqrt{2}} s i n (2 x - a) + c$ then a =

\frac{5 π}{4}

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- \frac{5 π}{4}

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\frac{π}{4}

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\frac{3 π}{4}

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Solution

The correct option is B $- \frac{5 π}{4}$
$\int (s i n 2 x - c o s 2 x) d x = - \frac{1}{2} [c o s 2 x + s i n 2 x] + c = - \frac{1}{2} \sqrt{2} . s i n (2 x + \frac{π}{4}) + c = - \frac{1}{\sqrt{2}} s i n (2 x + \frac{π}{4}) = \frac{1}{\sqrt{2}} s i n (2 x + \frac{5 π}{4}) \Rightarrow α = - \frac{5 π}{4}$

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Similar questions

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$\int (s i n 2 x - c o s 2 x) d x = \frac{1}{\sqrt{2}} s i n (2 x - a) + c$ then a =

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∫(sin 2x−cos 2x) dx=1√2sin(2x−a)+c then a =

The correct option is B −5π4∫(sin 2x−cos 2x) dx=−12[cos 2x+sin 2x]+c=−12√2.sin(2x+π4)+c=−1√2sin(2x+π4)=1√2 sin(2x+5π4)⇒α=−5π4

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$\int (s i n 2 x - c o s 2 x) d x = \frac{1}{\sqrt{2}} s i n (2 x - a) + c$ then a =