Step 1: Substitution
Let $v = x - y - 1$. Then $\dfrac{dv}{dx} = 1 - \dfrac{dy}{dx}$.
Step 2: Substitute $\dfrac{dy}{dx}$
Given $\dfrac{dy}{dx} = \cos^2(v)$, so:
$\dfrac{dv}{dx} = 1 - \cos^2(v) = \sin^2(v)$.
Step 3: Separate and integrate
$\dfrac{dv}{\sin^2(v)} = dx \Rightarrow \int \csc^2(v) \, dv = \int dx$
$- \cot(v) = x + C_1$
Step 4: Substitute back $v$
$- \cot(x - y - 1) = x + C_1$
Step 5: Solve for $x$
$x = - \cot(x - y - 1) - C_1$
Let $C = -C_1$, then:
$x = C - \cot(x - y - 1)$
The range of the real valued function \( f(x) =\) \(\sin^{-1} \left( \frac{1 + x^2}{2x} \right)\) \(+ \cos^{-1} \left( \frac{2x}{1 + x^2} \right)\) is:
If \(3A = \begin{bmatrix} 1 & 2 & 2 \\[0.3em] 2 & 1 & -2 \\[0.3em] a & 2 & b \end{bmatrix}\) and \(AA^T = I\), then\(\frac{a}{b} + \frac{b}{a} =\):
\(\begin{vmatrix} a+b+2c & a & b \\[0.3em] c & b+c+2c & b \\[0.3em] c & a & c+a2b \end{vmatrix}\)