int {{e^{2x}}left( {frac{{sin 4x - 2}}{{1 - c | vedclass

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(A) Let $I = \int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx$. Using the identities $\sin 4x = 2 \sin 2x \cos 2x$ and $1 - \c

Vedclass · Accepted Answer

$\frac{1}{2}{e^{2x}}\cot 2x + c$

Vedclass · Answer

(A) Let $I = \int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx$. Using the identities $\sin 4x = 2 \sin 2x \cos 2x$ and $1 - \cos 4x = 2 \sin^2 2x$,we get: $I = \int {{e^{2x}}\left( {\frac{{2 \sin 2x \cos 2x - 2}}{{2 \sin^2 2x}}} \right)} \,dx$ $I = \int {{e^{2x}}\left( {\frac{{\sin 2x \cos 2x - 1}}{{\sin^2 2x}}} \right)} \,dx$ $I = \int {{e^{2x}}\left( {\cot 2x - \csc^2 2x} \right)} \,dx$ Let $f(x) = \cot 2x$. Then $f'(x) = -2 \csc^2 2x$. This does not fit the form $\int e^{ax}(f(x) + f'(x)/a) dx$ directly. Let's integrate by parts: $\int e^{2x} \cot 2x \, dx - \int e^{2x} \csc^2 2x \, dx$. $int e^{2x} \cot 2x \, dx = \frac{1}{2} e^{2x} \cot 2x - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx$. Substituting this back: $I = \left( \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx \right) - \int e^{2x} \csc^2 2x \, dx + c$ $I = \frac{1}{2} e^{2x} \cot 2x + c$.

$\int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)\;dx = } $

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