int e^{2x} frac{(sin 2x cos 2x - 1)}{sin^2 2x | vedclass

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(D) Let $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$. We can rewrite the integrand as: $I = \int e^{2x} \left( \frac{\sin 2x \cos 2x}

Vedclass · Accepted Answer

$\frac{1}{2} e^{2x} \cot(2x) + c$,where $c$ is the constant of integration

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(D) Let $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$. We can rewrite the integrand as: $I = \int e^{2x} \left( \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} \right) \, dx$ $I = \int e^{2x} (\cot 2x - \csc^2 2x) \, dx$. Let $f(x) = \cot 2x$. Then $f'(x) = -\csc^2 2x \cdot 2 = -2 \csc^2 2x$. This does not fit the form $\int e^{ax} (f(x) + \frac{f'(x)}{a}) \, dx$. Let us rewrite the integral as $I = \int e^{2x} \cot 2x \, dx - \int e^{2x} \csc^2 2x \, dx$. Using integration by parts on $\int e^{2x} \cot 2x \, dx$: Let $u = \cot 2x$,$dv = e^{2x} \, dx$. Then $du = -2 \csc^2 2x \, dx$ and $v = \frac{1}{2} e^{2x}$. $I = (\cot 2x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx) - \int e^{2x} \csc^2 2x \, dx$. $I = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx - \int e^{2x} \csc^2 2x \, dx + c$. $I = \frac{1}{2} e^{2x} \cot 2x + c$.

$\int e^{2x} \frac{(\sin 2x \cos 2x - 1)}{\sin^2 2x} \, dx =$

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