Найти интеграл помогите
Чтобы вычислить такой интеграл, надо хорошо знать производные элементарных функций. Для этого примера надо помнить, что .
.![\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{cosx}{sin^2x}\, dx=\int\limits^{\frac{\pi }{2}}_{\frac{\pi}{4}}(sinx)^{-2}\cdot cosx\, dx=[\; t=sinx\; ,\; dt=cosx\, dx\; ,\\\\ t_1=sin\frac{\pi}{4}=\frac{\sqrt2}{2}\; ,\; t_2=sin\frac{\pi}{2}=1\; ]=\int\limits^1_{\frac{\sqrt2}{2}}t^{-2}\cdot dt=\frac{t^{-1}}{-1}\Big |_{\frac{\sqrt2}{2}}^1=-\frac{1}{t}\Big |^1_ {\frac{\sqrt2}{2}}=\\\\=-\frac{1}{1}+\frac{1}{\sqrt2/2}=\frac{2}{\sqrt2}-1=\sqrt2-1](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5Cfrac%7Bcosx%7D%7Bsin%5E2x%7D%5C%2C%20dx%3D%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%28sinx%29%5E%7B-2%7D%5Ccdot%20cosx%5C%2C%20dx%3D%5B%5C%3B%20t%3Dsinx%5C%3B%20%2C%5C%3B%20dt%3Dcosx%5C%2C%20dx%5C%3B%20%2C%5C%5C%5C%5C%20t_1%3Dsin%5Cfrac%7B%5Cpi%7D%7B4%7D%3D%5Cfrac%7B%5Csqrt2%7D%7B2%7D%5C%3B%20%2C%5C%3B%20t_2%3Dsin%5Cfrac%7B%5Cpi%7D%7B2%7D%3D1%5C%3B%20%5D%3D%5Cint%5Climits%5E1_%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7D%7Dt%5E%7B-2%7D%5Ccdot%20dt%3D%5Cfrac%7Bt%5E%7B-1%7D%7D%7B-1%7D%5CBig%20%7C_%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7D%7D%5E1%3D-%5Cfrac%7B1%7D%7Bt%7D%5CBig%20%7C%5E1_%20%7B%5Cfrac%7B%5Csqrt2%7D%7B2%7D%7D%3D%5C%5C%5C%5C%3D-%5Cfrac%7B1%7D%7B1%7D%2B%5Cfrac%7B1%7D%7B%5Csqrt2%2F2%7D%3D%5Cfrac%7B2%7D%7B%5Csqrt2%7D-1%3D%5Csqrt2-1 "\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{cosx}{sin^2x}\, dx=\int\limits^{\frac{\pi }{2}}_{\frac{\pi}{4}}(sinx)^{-2}\cdot cosx\, dx=[\; t=sinx\; ,\; dt=cosx\, dx\; ,\\\\ t_1=sin\frac{\pi}{4}=\frac{\sqrt2}{2}\; ,\; t_2=sin\frac{\pi}{2}=1\; ]=\int\limits^1_{\frac{\sqrt2}{2}}t^{-2}\cdot dt=\frac{t^{-1}}{-1}\Big |_{\frac{\sqrt2}{2}}^1=-\frac{1}{t}\Big |^1_ {\frac{\sqrt2}{2}}=\\\\=-\frac{1}{1}+\frac{1}{\sqrt2/2}=\frac{2}{\sqrt2}-1=\sqrt2-1")