If int frac{cos x-sin x}{sqrt{8-sin 2 x}} d x | vedclass

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(C) Given integral: $I = \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin

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$(1, 3)$

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(C) Given integral: $I = \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2\sin x \cos x) = 1 - (\sin x - \cos x)^2$. However,it is more convenient to write $\sin 2x = (\sin x + \cos x)^2 - 1$.Substituting this into the denominator:$8 - \sin 2x = 8 - ((\sin x + \cos x)^2 - 1) = 8 - (\sin x + \cos x)^2 + 1 = 9 - (\sin x + \cos x)^2$.Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.The integral becomes:$I = \int \frac{dt}{\sqrt{9 - t^2}} = \sin^{-1}\left(\frac{t}{3}\right) + c$.Substituting $t$ back:$I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$.Comparing this with $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we get $a = 1$ and $b = 3$.Thus,the ordered pair $(a, b)$ is $(1, 3)$.

If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$ where $c$ is a constant of integration,then the ordered pair $(a, b)$ is equal to

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