Помогите решить интеграл. Created by lepme. algebra-ru

Помогите решить интеграл...

lepme @lepme

August 2022 1 12 Report

Помогите решить интеграл

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$\int \frac{dx}{sin^2x\cdot cosx}=\int \frac{sin^2x+cos^2x}{sin^2x\cdot cosx}\, dx=\int \frac{sin^2x\, dx}{y\sin^2x\cdot cosx}+\int \frac{cos^2x\, dx}{sin^2x\cdot cosx}=\\\\=\int \frac{dx}{cosx}+\int \frac{cosx\, dx}{sin^2x}=ln|tg(\frac{x}{2}+\frac{\pi}{4})|+\int \frac{d(sinx)}{sin^2x}=ln|tg(\frac{x}{2}+\frac{\pi}{4})|-\frac{1}{sinx}+C\; ;\\\\\\\star \; \; \int \frac{dx}{cosx}=\int \frac{cosx\, dx}{cos^2x}=\int\frac{cosx\, dx}{1-sin^2x}=\int \frac{d(sinx)}{1-sin^2x}=\Big [\; \int \frac{dt}{1-t^2}=\frac{1}{2}\cdot ln|\frac{1+t}{1-t}|+C\; \Big ]=$

$=\frac{1}{2}\cdot ln\Big |\frac{1+sinx}{1-sinx}\Big |+C=\frac{1}{2}\cdot ln\Big |\frac{1+cos(\frac{\pi}{4}-\frac{x}{2})}{1-cos(\frac{\pi}{4}-\frac{x}{2})}\Big |+C=\\\\=\Big [\; 1+cos\alpha =2cos^2\frac{\alpha }{2}\; ,\; 1-cos\alpha =2sin^2\frac{\alpha }{2}\; \Big ]=\\\\=\frac{1}{2}\cdot ln\Big |\frac{2cos^2(\frac{\pi}{4}-\frac{x}{2})}{2sin^2(\frac{\pi}{4}-\frac{x}{2})}\Big |+C=ln\Big |ctg(\frac{\pi}{4}-\frac{x}{2})\Big |+C=\Big [\; ctg\alpha =tg(\frac{\pi}{2}-\alpha )\; \Big ]=\\\\=ln\Big |tg(\frac{\pi}{2}-(\frac{\pi}{4}-\frac{x}{2}))\Big |+C=ln\Big |tg(\frac{\pi}{4}+\frac{x}{2})\Big |+C\; .$

P.S. Желательно запомнить, кроме таблицы интегралов, ещё два часто встречающихся интеграла:

$\int \frac{dx}{sinx}=ln\Big |tg\frac{x}{2}\Big |+C\; ;\; \; \int \frac{dx}{cosx}=ln\Big |tg(\frac{x}{2}+\frac{\pi}{4})\Big |+C\; .$

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