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Solve the differential equation

$displaystyle frac{dy}{dx} = sqrt{frac{1}{2}+int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}},$

where $theta = x+y$ and $displaystyle x+yin bigg(frac{pi}{4};,frac{3pi}{4}bigg)$

Try: First we will sove $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots cdots (1)$$

Replace $phi = cos^4 theta-sin^4 theta-phi = cos (2theta)-phi.$

So $$I = int^{-sin^4
theta}_{cos^4 theta}frac{sqrt{f(cos 2 theta -phi)}dphi}{sqrt{f(theta)}+sqrt{f(cos 2 theta-phi)}}cdots (2)$$

So $$2I = int^{-sin^4 theta}_{cos^4 theta}dphi = phibigg|^{-sin^4 theta}_{cos^4 theta}$$

$$I = frac{1}{2}bigg[sin^4 theta +cos ^4 thetabigg] = frac{1}{2}-sin^2 theta cos^2 theta$$

So $$frac{dy}{dx} = sqrt{1-sin^2 theta cos^2 theta}$$

could some help me how to solve further, i don,t get any clue

thanks

answer given as $displaystyle frac{2}{sqrt{3}}tan^{-1}bigg(frac{2tan(x+y)+1}{sqrt{3}}bigg)=x+c.$

I'll assume that you meant to have $f(phi)$ and not $f(theta)$ in the denominator of the integral

You made a sign error

$$ I = frac12 left(-sin^4 theta - cos^4 theta right) = -frac12 +sin^2thetacos^2theta $$

Then
$$ frac{dy}{dx} = sqrt{frac12 + I} = bigvertsinthetacosthetavert $$

You also have

$$ frac{dtheta}{dx} = 1 + frac{dy}{dx} = 1 + bigvertsinthetacosthetabigvert
= begin{cases}
1 + sinthetacostheta, & theta in left(frac{pi}{4}, frac{pi}{2} right) \
1 - sinthetacostheta, & theta in left( frac{pi}{2}, frac{3pi}{4} right)
end{cases} $$

Separating leads to the integral

$$ int frac{1}{1+sinthetacostheta}dtheta = int frac{sec^2theta}{sec^2theta+tantheta}dtheta = intfrac{1}{u^2+u+1}du $$

where $u = tantheta$. Completing the square on the denominator gives

$$ intfrac{1}{u^2+u+1}du = intfrac{1}{left(u+frac12right)^2+frac34}du = frac{2}{sqrt{3}}arctanleft(frac{2u+1}{sqrt{3}}right) + c $$