 # y'=\cos^{2}x \cos y Jessie Lee 2021-12-27 Answered
${y}^{\prime }={\mathrm{cos}}^{2}x\mathrm{cos}y$
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Given, ${y}^{\prime }={\mathrm{cos}}^{2}x\cdot \mathrm{cos}y$
$⇒\frac{dy}{dx}=\mathrm{cos}2x\cdot \mathrm{cos}y$
$⇒\frac{dy}{\mathrm{cos}y}={\mathrm{cos}}^{2}x\cdot dx$
Integrate both side, we get

$⇒\int \mathrm{sec}y\cdot dy=\int \left(\frac{1+\mathrm{cos}2x}{2}\right)dx$
$⇒\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}y\right)=\frac{1}{2}\left(x+\frac{\mathrm{sin}2x}{2}\right)+c$
$⇒\mathrm{ln}\left(\mathrm{sec}y+\mathrm{tan}x\right)=\frac{x}{2}+\frac{\mathrm{sin}2x}{4}+c$
Hence, solution of above differential equation is:
$\mathrm{ln}\left(\mathrm{sec}y+\mathrm{tan}y\right)=\frac{x}{2}+\frac{\mathrm{sin}2x}{4}+c$

Let, $v=\mathrm{sec}y+\mathrm{tan}y$
differentiate both side, we get
$dv=\left(\mathrm{sec}y\cdot \mathrm{tan}y+\mathrm{sec}2y\right)dy$
$\int \frac{1}{v}dv=\mathrm{ln}v+c$
$=\mathrm{ln}\left(\mathrm{sec}y+\mathrm{tan}y\right)+c$
$\int \mathrm{sec}ydy=\mathrm{ln}\left(\mathrm{sec}y+\mathrm{tan}y\right)+c$

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$\frac{dy}{dx}={\mathrm{cos}}^{2}x\cdot \mathrm{cos}y$
Separate the variable x and y both side.

Simplify
Integrate both side of equation

$\mathrm{ln}|\mathrm{tan}y+\mathrm{sec}y|=\int \frac{\left(1+\mathrm{cos}2x\right)}{2}dx$
$=\frac{1}{2}\int \left(1+\mathrm{cos}2x\right)dx$
$\mathrm{ln}|\mathrm{tan}y+\mathrm{sec}y|=\frac{1}{2}\left[x+\frac{\mathrm{sin}2x}{2}\right]+c$

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