Определен интеграл1

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Да се реши

[aifC]

[tex]\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{sin^{2}(x)}{sin(x)+cos(x)+1}dx;[/tex] Правим субституция на Вайерщрас:

[tex]\int \frac{4tg^{2} \left(\frac{x}{2}\right)}{(tg^{2} \left(\frac{x}{2}\right)+1) \left(\frac{1-tg^{2} \left(\frac{x}{2}\right)}{tg^{2} \left(\frac{x}{2}\right)+1} + \frac{2tg \left(\frac{x}{2}\right)}{tg^{2} \left(\frac{x}{2}\right)+1}\right)} \Rightarrow[/tex]

[tex]u = tg \left(\frac{x}{2} \right) \rightarrow dx = \frac{2}{sec^{2} \left(\frac{x}{2}\right)}, du = \frac{2}{u^{2}+1}; \Rightarrow 4\int \frac{u^{2}}{(u+1)(u^{2}+1)^{2}}du;[/tex]

[tex]\int \left(-\frac{u-1}{4(u^{2}-1)} + \frac{u-1}{2(u^{2}-1)^{2}} + \frac{1}{4(u+1)} \right)du = -\frac{1}{4}\int \frac{u-1}{u^{2}+1}du + \frac{1}{2}\int \frac{u-1}{(u^{2}+1)^{2}}du + \frac{1}{4}\int \frac{1}{u+1}du = -I_{1}+I_{2}+I_{3};[/tex]

[tex]I_{1} = \int \frac{u-1}{u^{2}+1}du = \int \left(\frac{u}{u^{2}+1} - \frac{1}{u^{2}+1}\right)du = \int \frac{u}{u^{2}+1}du - \int \frac{1}{u^{2}+1}du;[/tex]

[tex]\int \frac{u}{u^{2}+1}du; v=u^{2}+1 \rightarrow du = \frac{1}{2u}dv \Rightarrow \frac{1}{2} \int \frac{1}{v}dv = \frac{ln(v)}{2} = \frac{ln(u^{2}+1)}{2};[/tex]

[tex]\int \frac{1}{u^{2}+1}du = arctg(u);[/tex]

[tex]I_{2} = \int \frac{u-1}{(u^{2}+1)^{2}}du = \int \left(\frac{u}{(u^{2}+1)^{2}} - \frac{1}{(u^{2}+1)^{2}}\right)du = \int \frac{u}{(u^{2}+1)^{2}} du - \int\frac{1}{(u^{2}+1)^{2}}du[/tex]

[tex]\int \frac{u}{(u^{2}+1)^{2}} du; u=u^{2}+1 \rightarrow du = \frac{1}{2u}dv \Rightarrow \frac{1}{2(u^{2}+1)};[/tex]

[tex]\int\frac{1}{(u^{2}+1)^{2}} = \frac{1}{2} \int \frac{1}{u^{2}+1}du + \frac{u}{2(u^{2}+1)} \Rightarrow arctg(u);[/tex]

Обединяваме решенията връщаме субституцията и пресмятаме опрделения интеграл:

[tex]-\frac{sin(x)+cos(x)-2ln(|sin \left(\frac{x}{2}\right)+cos \left(\frac{x}{2}\right)}{2} \begin{array}{|l} \frac{\pi}{2} \\ \frac{\pi}{3}\end{array} - \left(-\frac{sin(x)+cos(x)-2ln(|sin \left(\frac{x}{2}\right)+cos \left(\frac{x}{2}\right)}{2} \begin{array}{|l} \frac{\pi}{2} \\ \frac{\pi}{3}\end{array}\right) =[/tex]

[tex]\frac{1}{2}(ln(2)-1)+\frac{1}{4} + \frac{\sqrt{3}}{4} + ln \left(\frac{1+\sqrt{3}}{2} \right) = ln \left(\frac{1+\sqrt{3}}{2} \right) + \frac{2ln(2)+\sqrt{3}-1}{2}[/tex]

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Благодаря.