Analiza matematyczna, zadanie nr 1836

Autor	Zadanie / Rozwiązanie
pm12 postów: 493	2013-12-26 14:40:34 Całki nieoznaczone : 1. $\frac{1+x^{2}}{1+x^{4}}$ 2. $\frac{1}{cosx}$ 3. $\frac{1}{sin^{2}xcos4x}$

abcdefgh postów: 1255	2013-12-26 16:08:35 $\int \frac{1}{cosx}=\begin{bmatrix}t=tg(\frac{x}{2})\\dx=\frac{2}{1+t^2}dt\\cosx=\frac{1-t^2}{1+t^2}\end{bmatrix}=\int \frac{1}{\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=\int \frac{2}{1-t^2}dt=2\int \frac{1}{(1-t)(1+t)dt}=2\frac{1}{2}(\int \frac{1}{1-t}dt+\int \frac{1}{1+t}dt$ $int \frac{1}{1-t}dt+\int \frac{1}{1+t}dt=-ln\|1-t\|+ln\|1+t\|=ln\|\frac{1-t}{1+t}\|=ln\|\frac{1-tg(\frac{x}{2}}{1+tg(\frac{x}{2}}\|$

abcdefgh postów: 1255	2013-12-26 18:01:17 $\int \frac{1+x^{2}}{1+x^{4}}$ $\frac{1+x^{2}}{1+x^{4}}=\frac{1}{2}(\frac{1}{(x^2+\sqrt{2}x+1)}+\frac{1}{(x^2-\sqrt{2}x+1)})$ $\frac{1}{2}(\int \frac{1}{(x^2+\sqrt{2}x+1)} +\int\frac{1}{(x^2-\sqrt{2}x+1)})=$ 1)$\int \frac{1}{(x^2+\sqrt{2}x+1)}=\int \frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=2\int \frac{1}{(\frac{x+\frac{\sqrt{2}}{2}}{\frac{1}{\sqrt{2}}})^2+1}=2\int \frac{1}{(\sqrt{2}x+1)^2+1}=\begin{bmatrix}t=\sqrt{2}x+1\\dt=\sqrt{2}dx\\dx=\frac{dt}{\sqrt{2}}\end{bmatrix}$ $=\frac{2}{\sqrt{2}}\int \frac{1}{t^2+1}=\frac{2}{\sqrt{2}}arctg(\sqrt{2}+1)$ 2)$\int \frac{1}{(x^2-\sqrt{2}x+1)}=\int \frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}=2\int \frac{1}{(\frac{x-\frac{\sqrt{2}}{2}}{\frac{1}{\sqrt{2}}})^2+1}=2\int \frac{1}{(\sqrt{2}x-1)^2+1}=\begin{bmatrix}t=\sqrt{2}x-1\\dt=\sqrt{2}dx\\dx=\frac{dt}{\sqrt{2}}\end{bmatrix}$ $=\frac{2}{\sqrt{2}}\int \frac{1}{t^2+1}=\frac{2}{\sqrt{2}}arctg(\sqrt{2}-1)$ $=\frac{2}{2\sqrt{2}}arctg(\sqrt{2}-1)+\frac{2}{2\sqrt{2}}arctg(\sqrt{2}+1)$

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