If int (cos x - sin x) , dx = sqrt{2} sin (x | vedclass

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Question

(C) Given that $\int (\cos x - \sin x) \, dx = \sqrt{2} \sin (x + \alpha) + c$. Integrating the left side: $\int (\cos x - \sin x) \, dx = \sin x + \c

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given that $\int (\cos x - \sin x) \, dx = \sqrt{2} \sin (x + \alpha) + c$. Integrating the left side: $\int (\cos x - \sin x) \, dx = \sin x + \cos x + c$. Now,equate this to the right side: $\sin x + \cos x + c = \sqrt{2} \sin (x + \alpha) + c$. Divide and multiply the left side by $\sqrt{2}$: $\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) + c = \sqrt{2} \sin (x + \alpha) + c$. Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. So,$\sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) + c = \sqrt{2} \sin (x + \alpha) + c$. This simplifies to $\sqrt{2} \sin (x + \frac{\pi}{4}) + c = \sqrt{2} \sin (x + \alpha) + c$. Comparing both sides,we get $\alpha = \frac{\pi}{4}$.

If $\int (\cos x - \sin x) \, dx = \sqrt{2} \sin (x + \alpha) + c$,then $\alpha = $

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