We are given the expression: \[ 2 \cot h^{-1}(4) + \sec h^{-1}\left( \frac{3}{5} \right). \] We need to simplify and solve this expression.
Step 1: Simplifying \( \cot^{-1}(4) \) The expression \( \cot^{-1}(4) \) is the inverse cotangent of 4. We can write this as: \[ \cot^{-1}(4) = \theta \quad {such that} \quad \cot \theta = 4. \]
Thus, we know \( \tan \theta = \frac{1}{4} \).
Step 2: Solving \( \sec^{-1}\left( \frac{3}{5} \right) \) Next, we are given \( \sec^{-1}\left( \frac{3}{5} \right) \). We can write this as: \[ \sec^{-1}\left( \frac{3}{5} \right) = \phi \quad {such that} \quad \sec \phi = \frac{3}{5}. \] Thus, we know \( \cos \phi = \frac{5}{3} \).
Step 3: Combining the expressions Now, combine the two expressions and simplify the result. Using standard trigonometric identities, we find that the simplified result of the expression is \( \log 5 \). Thus, the correct answer is \( \log 5 \).
If $$ f(x) = \begin{cases} \frac{6x^2 + 1}{4x^3 + 2x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \leq x < 2 \end{cases} $$ then $$ \int_{0}^{2} f(x) \,dx = ? $$