Integrating function defined in terms of itself

Question

Here's the question I am trying to solve:

The variables $x$ and $y$ satisfy the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x} = 4\cos^2{y}\tan{x}.$$ for $0 \leq x < \frac{1}{2}\pi$, and $x = 0$ when $y = \frac{1}{4}\pi$. Solve this differential equation and find the value of $x$ when $y = \frac{1}{3}\pi$.

The answer (from the book) is:

$\tan{y} = 4\ln{\sec{x}} + 1$

I started by integrating both sides, to give $$y = \int{4\cos^2{y}\tan{x}\:\mathrm{d}x}.$$ But here I am stuck. I have no idea how to get rid of the $y$ in $4\cos^2{y}\tan{x}$. Normally with questions which have a $y$ on both sides, I can manipulate it into the form of $$\int{\frac{y\prime}{y}\:\mathrm{d}x} = \int{u}$$ which then simplifies to $$\ln{\left|y\right|} = \int{u}.$$

For this question I don't see how I can do this, as the $y$ is wrapped in a $\cos$ and a $a \to a ^ 2$.

Answer 1

What if you instead divide both sides by $ \cos^2 y $ (assume this is nonzero), producing

$$ \frac{1}{\cos^2y} \, \frac{dy}{dx} = 4 \tan x $$

and integrate both sides? You will then have to compute two integrals, one for each side of the equation:

$$ \int \frac{1}{\cos^2 y} \, dy = \int 4 \tan x \, dx. $$

Remember to include a constant (only one is needed), say $ C $, after integrating and use the initial conditions to find the value of $C$.

This is an example of a separable differential equation, where the general method is to isolate all the $x$s and $y$s to different sides and integrate.

Answer 2

$$\frac{dy}{dx}=4\cos^2(y)\tan(x)$$ so: $$\sec^2(y)\frac{dy}{dx}=4\tan(x)$$ now integrate both sides: $$\int\sec^2(y)\frac{dy}{dx}dx=4\int\tan(x)dx$$ $$\to$$ $$\int\sec^2(y)dy=4\int\tan(x)dx$$ $$\tan(y)=\ln|\sec(x)|+C$$ and we know that when $x=0$,$y=\frac{\pi}{4}$

so: $$\tan\left(\frac{\pi}{4}\right)=\ln(\sec(0))+C$$ $$1=0+C \therefore C=1$$ input this and we get: $$\tan(y)=\ln|\sec(x)|+1$$

Answer 3

$$f'(x)=a\cos^2(f(x))\tan(x)$$ $$\frac{f'(x)}{\cos^2(f(x))}=a\tan x$$ $$\int\frac{f'(x)}{\cos^2(f(x))}dx=a\int\tan(x)dx$$ $$\int\frac{f'(x)}{\cos^2(f(x))}dx=c_1-a\ln|\cos x|$$ Let $y=f(x)$. Therefore $dy=f'(x)dx$. Which gives $$\int\frac{dy}{\cos^2(y)}=c_1-a\ln|\cos x|$$ $$\int\sec^2y\,dy=c_1-a\ln|\cos x|$$ $$\tan y=c_1-a\ln|\cos x|$$ $$\tan f(x)=c_1-a\ln|\cos x|$$ Therefore $$f(x)=\arctan(c_1-a\ln|\cos x|)$$