Let the slope of the tangent to a curve y=f(x | vedclass

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(B) The slope of the tangent is given by $\frac{dy}{dx} = 2 	an x(\cos x - y)$.This can be rewritten as the linear differential equation: $\frac{dy}{

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$2-\frac{\pi}{\sqrt{2}}$

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(B) The slope of the tangent is given by $\frac{dy}{dx} = 2 	an x(\cos x - y)$.This can be rewritten as the linear differential equation: $\frac{dy}{dx} + (2 	an x)y = 2 \sin x$.The integrating factor is $IF = e^{\int 2 	an x \, dx} = e^{2 \ln |\sec x|} = \sec^2 x$.Multiplying both sides by $IF$,we get $\frac{d}{dx}(y \sec^2 x) = 2 \sin x \sec^2 x = 2 \sec x 	an x$.Integrating both sides,$y \sec^2 x = 2 \sec x + C$,which simplifies to $y = 2 \cos x + C \cos^2 x$.Since the curve passes through $(\frac{\pi}{4}, 0)$,we have $0 = 2 \cos(\frac{\pi}{4}) + C \cos^2(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) + C(\frac{1}{2}) = \sqrt{2} + \frac{C}{2}$.Thus,$C = -2\sqrt{2}$,and the curve is $y = 2 \cos x - 2\sqrt{2} \cos^2 x$.Now,$\int_{0}^{\pi / 2} y \, dx = \int_{0}^{\pi / 2} (2 \cos x - 2\sqrt{2} \cos^2 x) \, dx$.$= [2 \sin x]_{0}^{\pi / 2} - 2\sqrt{2} \int_{0}^{\pi / 2} \frac{1 + \cos 2x}{2} \, dx$.$= 2(1 - 0) - \sqrt{2} [x + \frac{\sin 2x}{2}]_{0}^{\pi / 2} = 2 - \sqrt{2}(\frac{\pi}{2}) = 2 - \frac{\pi}{\sqrt{2}}$.

Let the slope of the tangent to a curve $y=f(x)$ at $(x, y)$ be given by $2 \tan x(\cos x-y)$. If the curve passes through the point $(\frac{\pi}{4}, 0)$,then the value of $\int_{0}^{\pi / 2} y \, dx$ is equal to

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