If int (sin 2x + cos 2x) dx = frac{1}{sqrt{2} | vedclass

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Question

(A) We are given the integral $\int (\sin 2x + \cos 2x) dx$. Integrating term by term,we get: $\int \sin 2x dx + \int \cos 2x dx = -\frac{\cos 2x}{2}

Vedclass · Accepted Answer

$c = \pi/4$ and $a = k$ (an arbitrary constant)

Vedclass · Answer

(A) We are given the integral $\int (\sin 2x + \cos 2x) dx$. Integrating term by term,we get: $\int \sin 2x dx + \int \cos 2x dx = -\frac{\cos 2x}{2} + \frac{\sin 2x}{2} + k$,where $k$ is an arbitrary constant. We can rewrite this expression as: $\frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} \sin 2x - \frac{1}{\sqrt{2}} \cos 2x \right) + k$. Since $\cos(\pi/4) = 1/\sqrt{2}$ and $\sin(\pi/4) = 1/\sqrt{2}$,the expression becomes: $\frac{1}{\sqrt{2}} (\sin 2x \cos(\pi/4) - \cos 2x \sin(\pi/4)) + k$. Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get: $\frac{1}{\sqrt{2}} \sin(2x - \pi/4) + k$. Comparing this with the given form $\frac{1}{\sqrt{2}} \sin(2x - c) + a$,we find $c = \pi/4$ and $a = k$ (an arbitrary constant).

If $\int (\sin 2x + \cos 2x) dx = \frac{1}{\sqrt{2}} \sin (2x - c) + a$,then the value of $a$ and $c$ is:

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