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

Option: 1

\mathrm{(2-\sqrt{2})+\frac{\pi}{\sqrt{2}}}


Option: 2

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


Option: 3

(2+\sqrt{2})+\frac{\pi}{\sqrt{2}}


Option: 4

\mathrm{2+\frac{\pi}{\sqrt{2}} }


Answers (1)

best_answer

\mathrm{\frac{d y}{d x}=2 \tan x(\cos x-y) } \\

\mathrm{\Rightarrow \frac{d y}{d x}=2 \sin x-2 \tan x \cdot y} \\

\mathrm{\Rightarrow \frac{d y}{d x}+2 \tan x \cdot y=2 \sin x }\\

\mathrm{\text { I.F. } =e^{\int 2 \tan x d x}=e^{\ln \sec ^{2} x}=\sec ^{2} x}

\mathrm{\Rightarrow y \cdot \sec ^{2} x=2 \int \sin x.\sec ^{2} x d x }\\

\mathrm{\Rightarrow y \cdot \sec ^{2} x=2 \int \tan x \cdot \sec x d x }\\

\mathrm{\Rightarrow y \cdot \sec ^{2} x=2 \sec x+c}

\mathrm{Putting \left(\frac{\pi}{4}, 0\right)}

\mathrm{0=2 \sqrt{2}+c \Rightarrow c=-2 \sqrt{2}} \\

\mathrm{\Rightarrow y \cdot \sec ^{2} x=2 \sec x-2 \sqrt{2}} \\

\mathrm{\Rightarrow y=2 \cos x-2 \sqrt{2} \cos ^{2} x }\\

\mathrm{\Rightarrow y=2 \cos x-\sqrt{2}(1+\cos 2 x) }

\mathrm{\therefore \quad \int_{0}^{\pi / 2} y d x=\int_{0}^{\pi / 2} 2 \cos x d x-\sqrt{2} \int_{0}^{\pi / 2}(1+\cos 2 x) d x }\\

\mathrm{=\left.2 \sin x\right|_{0} ^{\pi / 2}-\left.\sqrt{2}\left(x+\frac{\sin 2 x}{x}\right)\right|_{0} ^{\pi / 2}}

Hence the correct answer is option 2.

Posted by

rishi.raj

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