- sin 2x-cos 2x dx - 1-2sin2x-a-c then a -

∫ (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 = 5 π/4 ...

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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

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Similar questions

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

\int (sin 2 x - cos 2 x) d x = \frac{1}{\sqrt{2}} sin (2 x - a) + C

\int \frac{{sin}^{8} x - {cos}^{8} x}{1 - 2 {sin}^{2} x {cos}^{2} x} d x = A sin 2 x + B

A =

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

\frac{d y}{d x}

y = \frac{1 + sin 2 x - cos 2 x}{1 + sin 2 x + cos 2 x}

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