# Solve for G.S. / P.S. for the following differential equations using separation of variables. 2 sin^2 2t dx=cos x(1+cos 2x)dt

Solve for G.S. / P.S. for the following differential equations using separation of variables.
$2{\mathrm{sin}}^{2}2tdx=\mathrm{cos}x\left(1+\mathrm{cos}2x\right)dt$
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Step 1
the given differential equation is:
$2{\mathrm{sin}}^{2}2tdx=\mathrm{cos}x\left(1+\mathrm{cos}2x\right)dt$
we have to find the general solution of the given differential equation by using separation of variables.
separate the variables:
$2{\mathrm{sin}}^{2}2tdx=\mathrm{cos}x\left(1+\mathrm{cos}2x\right)dt$
$\frac{dx}{\mathrm{cos}x\left(1+\mathrm{cos}2x\right)}=\frac{dt}{2{\mathrm{sin}}^{2}2t}$
now integrate the both sides of the equation.
Therefore,
$\int \frac{dx}{\mathrm{cos}x\left(1+\mathrm{cos}2x\right)}=\int \frac{dt}{2{\mathrm{sin}}^{2}2t}\left(1\right)$
Step 2
let the integral $\int \frac{dx}{\mathrm{cos}x\left(1+\mathrm{cos}2x\right)}$
therefore,
$I=\int \frac{dx}{\mathrm{cos}x\left(1+\mathrm{cos}2x\right)}$
now as we know that $1+\mathrm{cos}2x=2{\mathrm{cos}}^{2}x$
therefore,
$I=\int \frac{dx}{\mathrm{cos}x\left(2{\mathrm{cos}}^{2}x\right)}$
$=\frac{1}{2}\int \frac{dx}{{\mathrm{cos}}^{3}x}$
$=\frac{1}{2}\int {\mathrm{sec}}^{3}xdx$
step 3
now let
therefore,
${I}_{1}=\int {\mathrm{sec}}^{3}xdx$
$=\int {\mathrm{sec}}^{2}x\mathrm{sec}xdx$
now let ${\mathrm{sec}}^{2}x$ be the second function and the secx be the first function and solve the integral ${I}_{1}$ by using integration by parts. therefore,
${I}_{1}=\int {\mathrm{sec}}^{2}x\mathrm{sec}xdx$
${I}_{1}=\mathrm{sec}x\int {\mathrm{sec}}^{2}xdx-\int \left(\frac{d}{dx}\left(\mathrm{sec}x\right)\int {\mathrm{sec}}^{2}xdx\right)dx$