Монгол Бодлогын Сан

Заавар: $$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$ ашиглан хувьсагчийг ялга.

Бодолт: $$y^\prime+\cos(x+2y)=\cos(x-2y)\Leftrightarrow y^\prime=2\sin x\sin 2y$$ тул $$\dfrac{dy}{\sin 2y}=2\sin xdx\Leftrightarrow\dfrac{d\sin y}{2\sin y(1-\sin^2y)}=2\sin xdx$$ болно. Нөгөө талаас \begin{align*} \dfrac{1}{x(1-x^2)}&=\dfrac{-1}{(x-1)x(x+1)}\\ &=\left(\dfrac{1}{x}-\dfrac{1}{x-1}\right)\dfrac{1}{x+1}\\ &=\dfrac{1}{x(x+1)}-\dfrac{1}{(x-1)(x+1)}\\ &=\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)}\\ &=-\dfrac{1}{2(x-1)}+\dfrac{1}{x}-\dfrac{1}{2(x+1)} \end{align*} буюу $$\displaystyle\int\dfrac{1}{x(1-x^2)}=-\frac12\ln|x-1|+\ln |x|-\frac12\ln|x+1|+C=\frac12\ln\left|\dfrac{x^2}{x^2-1}\right|+C$$ тул $$\dfrac12\ln\left|\dfrac{\sin^2 y}{\sin^2y-1}\right|=-2\cos x+C$$ $x=0$ үед $y=\dfrac{\pi}{4}$ тул $$\dfrac12\ln\left|\dfrac{\left(\frac{\sqrt2}{2}\right)^2}{\left(\frac{\sqrt2}{2}\right)^2-1}\right|=-2\cos 0+C\Rightarrow C=2$$ Түүнчлэн $\left|\dfrac{\sin^2y}{\sin^2y-1}\right|=\tg^2y$ тул $$\tg y=e^{2-2\cos x}$$ байна.