If the integral of (sin 2x - cos 2x) dx = 1/√2(sin(2x - a) + b), then find a & b

Solution:

The function can be solved as a set of two different equations, LHS and RHS separately.

∫(sin 2x - cos 2x) dx = 1/√2(sin(2x - a) + b)

LHS =  ∫(sin 2x - cos 2x) dx

RHS = 1/√2(sin(2x - a) + b)

LHS =  -1/2(cos 2x + sin 2x) + c

Multiplying and dividing the above by √2, we get,

LHS = -1/2(cos 2x + sin 2x)√2/√2 + c

LHS = 1/√2[cos 2x (-1/√2) + sin 2x (-1/√2)] + c

We know that sine and cosine are both negative in the third quadrant. Therefore, -1/√2 = sin 5π/4 = cos 5π/4 

LHS = 1/√2 [cos 2x sin (5π/4) + sin 2x cos (5π/4)] + c

LHS = 1/√2 sin [2x + (5π/4)] + c

RHS = 1/√2 (sin (2x - a) + b)

Comparing the above 2 equations, we can see that a = -(5π/4) and b = constant, b ∈ All real numbers

Therefore, a = -(5π/4) and b ∈ All real numbers. 

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If the integral of (sin 2x - cos 2x) dx = 1/√2(sin(2x - a) + b), then find a & b

Summary:

If the integral of (sin 2x - cos 2x) dx = 1/√2(sin(2x - a) + b) then a = -(5π/4) and b ∈ R.