Ребят, хелп, надо решить д.у. y`sin^2(x)=cos^2(x) y(pi/4)=(3pi)/4

Решите задачу:

$y'\cdot sin^2x=cos^2x\; ,\quad y( \frac{\pi }{4})= \frac{3\pi }{4} \\\\ \frac{dy}{dx}= \frac{cos^2x}{sin^2x}\\\\\int dy=\int ctg^2x\, dx\\\\\int dy=\int (1-\frac{1}{sin^2x})dx\\\\y= x+ctgx+C\\\\y( \frac{\pi }{4} )= \frac{\pi}{4}+ctg\frac{\pi }{4}+C\\\\ \frac{3\pi}{4}=\frac{\pi}{4}+1+C\\\\C= \frac{3\pi}{4}-\frac{\pi }{4}-1= \frac{\pi}{2}-1\\\\Otvet:\; \; y=x+ctgx+ \frac{\pi }{2} -1$