Evaluate the integral: \[ \int \frac{dx}{4 + 3\cot x} \]
\( \frac{4}{25} \log |4 + 3\cot x| - \frac{3}{25} x + C \)
Step 1: Substituting Trigonometric Identities We rewrite \( 4 + 3\cot x \) using the sine and cosine functions: \[ 4 + 3\cot x = 4 + 3 \frac{\cos x}{\sin x} \] Multiplying numerator and denominator by \( \sin x \), we obtain: \[ = \frac{4\sin x + 3\cos x}{\sin x} \] Thus, the given integral becomes: \[ I = \int \frac{\sin x \, dx}{4\sin x + 3\cos x} \]
Step 2: Using Substitution Let: \[ u = 4\sin x + 3\cos x \] Differentiating both sides: \[ du = (4\cos x - 3\sin x) dx \] We rewrite the integral using substitution: \[ I = \int \frac{dx}{4 + 3\cot x} \] Using standard integral properties: \[ I = -\frac{3}{25} \log |4\sin x + 3\cos x| + \frac{4}{25} x + C \]
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

