If int sqrt{2} sqrt{1 + sin x} , dx = -4 cos( | vedclass

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(A) Given integral: $I = \int \sqrt{2} \sqrt{1 + \sin x} \, dx$. Using the identity $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) =

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$\frac{1}{2}, \frac{\pi}{4}$

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(A) Given integral: $I = \int \sqrt{2} \sqrt{1 + \sin x} \, dx$. Using the identity $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\sin(x/2) + \cos(x/2))^2$. So,$I = \int \sqrt{2} (\sin(x/2) + \cos(x/2)) \, dx$. $I = \sqrt{2} \int \sin(x/2) \, dx + \sqrt{2} \int \cos(x/2) \, dx$. $I = \sqrt{2} [ -2 \cos(x/2) + 2 \sin(x/2) ] + c$. $I = 2\sqrt{2} [ \sin(x/2) - \cos(x/2) ] + c$. Multiplying and dividing by $\sqrt{2}$: $I = 2\sqrt{2} \cdot \sqrt{2} [ \frac{1}{\sqrt{2}} \sin(x/2) - \frac{1}{\sqrt{2}} \cos(x/2) ] + c$. $I = 4 [ \sin(x/2) \cos(\pi/4) - \cos(x/2) \sin(\pi/4) ] + c$. $I = 4 \sin(x/2 - \pi/4) + c$. Since $\sin(	heta) = -\cos(	heta + \pi/2)$,we have: $I = -4 \cos(x/2 - \pi/4 + \pi/2) + c = -4 \cos(x/2 + \pi/4) + c$. Comparing with $-4 \cos(ax + b) + c$,we get $a = 1/2$ and $b = \pi/4$.

If $\int \sqrt{2} \sqrt{1 + \sin x} \, dx = -4 \cos(ax + b) + c$,then the value of $(a, b)$ is

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