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Question Number 64687 by aliesam last updated on 20/Jul/19

Commented by mathmax by abdo last updated on 21/Jul/19

$$letA=\int \frac{e^{\frac{-cos\left(2x\right)}{2}}}{{sin}^{2}x}dx\Rightarrow A=\int \frac{e^{-\frac{cos\left(2x\right)}{2}}}{\frac{1-cos\left(2x\right)}{2}}dx$$$$=2\int \frac{e^{-\frac{cos\left(2x\right)}{2}}}{1-cos\left(2x\right)}dxchangementcos\left(2x\right)=tgive2x=argch\left(t\right)$$$$=ln(t+\sqrt{t^2-1})\Rightarrow x=\frac{1}{2}ln(t+\sqrt{t^2-1})\Rightarrow dx=\frac{1}{2}\frac{1+\frac{2t}{2\sqrt{t^2-1}}}{t+\sqrt{t^2-1}}$$$$=\frac{1}{2\sqrt{t^2-1}}\Rightarrow A=2\int \frac{e^{-\frac{t}{2}}}{1-t}\frac{dt}{2\sqrt{t^2-1}}$$$$=-\int \frac{e^{-\frac{t}{2}}}{(t-1)\sqrt{t^2-1}}dtcha7gementt=ch\left(u\right)give$$$$A=-\int \frac{e^{-\frac{chu}{2}}}{(chu-1)shu}sh\left(u\right)du=\int \frac{e^{-\frac{chu}{2}}}{1-ch\left(u\right)}du$$$$=\int \frac{e^{-\frac{chu}{2}}}{1-\frac{e^u+e^{-u}}{2}}du=\int \frac{2e^{-\frac{chu}{2}}}{2-e^u-e^{-u}}du$$$$=\int \frac{2e^{-u}{e}^{-\frac{chu}{2}}}{2e^{-u}-1-e^{-2u}}du...becontinued....$$$$$$

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