question_answer

Evaluate \[\int{{{e}^{x}}\left( \frac{\sin 4x-4}{1-\cos 4x} \right)}\,dx.\]

Answer:

Let \[l=\int{{{e}^{x}}}\left( \frac{\sin 4x-4}{1-\cos 4x} \right)\,\,dx=\int{{{e}^{x}}=\frac{2\sin 2x\cdot \cos 2x-4}{2{{\sin }^{2}}2x}\,dx}\]\[=\int{{{e}^{x}}=\frac{2\sin 2x\cdot \cos 2x}{2{{\sin }^{2}}2x}dx}=\int{{{e}^{x}}\frac{4}{2{{\sin }^{2}}2x}dx}\] \[=\int{{{e}^{x}}=\frac{\cos 2x}{\sin 2x}dx}-\int{{{e}^{x}}2\cos e{{c}^{2}}2xdx}\] \[=\int{{{e}^{x}}=\cot 2xdx-\int{2{{e}^{x}}}\cos e{{c}^{2}}2xdx}\] \[=\cot 2x\int{{{e}^{x}}dx-\int{\left\{ \frac{d\,(cot2x)}{dx}\int{{{e}^{x}}dx} \right\}dx}}\] \[-\int{2{{e}^{x}}\cos e{{c}^{2}}2xdx}\] [using integration by parts in I St integral] \[=\cot 2x\cdot {{e}^{x}}-\int{(-cose{{c}^{2}}2x)(2){{e}^{x}}dx-\int{2{{e}^{x}}\cos e{{c}^{2}}2xdx}}\]\[={{e}^{x}}\cot 2x+\int{2{{e}^{x}}\cos e{{c}^{2}}2xdx-\int{2{{e}^{x}}\cos e{{c}^{2}}2xdx}}\] \[={{e}^{x}}\cot 2x+C\]