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<p>\frac{d}{dx} [\sin(2x)]</p> <p>= \cos(2x) \cdot \frac{d}{dx}[2x]</p> <p>= \cos(2x) \cdot 2</p> <p>= 2\cos(2x)</p>

<p>\( \int f(x) \, dx = \int \frac{1}{3x} \, dx \)</p> <p>\( = \frac{1}{3} \int \frac{1}{x} \, dx \)</p> <p>\( = \frac{1}{3} \ln|x| + C \)</p> <p>Where \( C \) is the constant of integration.</p>

<p>\[\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C\]</p> <p>where \(C\) is the constant of integration.</p>

<p>\[ \int f(x) \, dx = \int x^{-2} \, dx \]</p> <p>\[ = \int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C \]</p> <p>\[ = \frac{x^{-1}}{-1} + C \]</p> <p>\[ = -\frac{1}{x} + C \]</p>

<p>\int 7^x dx = \int e^{x \ln(7)} dx</p> <p>Let u = x \ln(7) \Rightarrow du = \ln(7) dx</p> <p>dx = \frac{du}{\ln(7)}</p> <p>\int e^{x \ln(7)} dx = \int e^u \frac{du}{\ln(7)}</p> <p>\frac{1}{\ln(7)}\int e^u du = \frac{1}{\ln(7)} e^u + C</p> <p>\frac{1}{\ln(7)} e^{x \ln(7)} + C = \frac{7^x}{\ln(7)} + C</p> <p>So, the integral of f(x) = 7^x is \frac{7^x}{\ln(7)} + C</p>

<p>The integral of \( f(x) = -3 \cos(4x) \) can be found using the standard integration techniques for trigonometric functions.</p> <p>Let \( u = 4x \). Therefore, \( du = 4dx \) or \( \frac{du}{4} = dx \).</p> <p>The integral becomes:</p> <p>\( \int -3 \cos(4x) dx = \int -3 \cos(u) \frac{du}{4} \)</p> <p>\( = -\frac{3}{4} \int \cos(u) du \)</p> <p>\( = -\frac{3}{4} \sin(u) + C \)</p> <p>Substituting back \( u = 4x \):</p> <p>\( = -\frac{3}{4} \sin(4x) + C \)</p> <p>Where \( C \) is the constant of integration.</p>