## Całkowanie wyrażeń zawierających funkcje trygonometryczne

Uwaga 13.26.

Aby policzyć całkę

$$\displaystyle \int R(\sin x,\cos x,\mathrm{tg}\, x)\,dx,$$

stosujemy podstawienie

$$\displaystyle \mathrm{tg}\,\frac{x}{2}=t$$

i mamy

$$\begin{array}{rllll} \displaystyle \sin x & = & \displaystyle\frac{2\mathrm{tg}\,\frac{x}{2}}{1+\mathrm{tg}\,^2\frac{x}{2}} & = & \displaystyle\frac{2t}{1+t^2}, \\ \\ \ \cos x & = & \displaystyle\frac{1-\mathrm{tg}\,^2\frac{x}{2}}{1+\mathrm{tg}\,^2\frac{x}{2}} & = & \displaystyle\frac{1-t^2}{1+t^2}, \\ \\ \mathrm{tg}\, x & = & \displaystyle\frac{2\mathrm{tg}\,\frac{x}{2}}{1-\mathrm{tg}\,^2\frac{x}{2}} & = & \displaystyle\frac{2t}{1-t^2} \end{array}$$

oraz

$$\displaystyle x \ =\ 2\mathrm{arctg}\, t, \quad\$$ zatem $$\quad \,dx=\frac{2\,dt}{1+t^2}.$$ Po podstawieniu dostajemy całkę

$$\displaystyle \int R\bigg(\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}, \frac{2t}{1-t^2} \bigg)\frac{2dt}{1+t^2}.$$

Obliczyć całkę $$\displaystyle \int\frac{dx}{2+\cos x}.$$ W całce tej stosujemy podstawienie $$\displaystyle \mathrm{tg}\, \frac{x}{2}=t,$$ wówczas $$\displaystyle x=2\mathrm{arctg}\, t$$ i $$\displaystyle dx=\frac{2\,dt}{1+t^2}.$$ Zatem

$$\begin{array}{lll} \displaystyle \int\frac{dx}{2+\cos x} & = & \int\frac{\displaystyle\frac{2\,dt}{1+t^2}}{\displaystyle 2+\frac{1-t^2}{1+t^2}} \ =\ 2\int \frac{dt}{t^2+3} \ =\ \frac{2}{3}\int\frac{dt}{\displaystyle\bigg(\frac{t}{\sqrt{3}}\bigg)^2+1} \ =\ \bigg| \begin{array} {rcl} \displaystyle\frac{t}{\sqrt{3}} & = & s \\ dt & = & \displaystyle\sqrt{3}\,ds \end{array} \bigg| \\ & = & \displaystyle \frac{2}{\sqrt{3}}\int\frac{\displaystyle\sqrt{3}\,ds}{s^2+1} \ =\ 2\mathrm{arctg}\, s+c \ =\ 2\mathrm{arctg}\,\bigg(\displaystyle\frac{\mathrm{tg}\,\frac{x}{2}}{\sqrt{3}}\bigg)+c. \end{array}$$

Uwaga 13.28.

Aby policzyć całkę

$$\displaystyle \int R(\sin^2x,\cos^2x,\sin x\cos x)\,dx,$$

stosujemy podstawienie

$$\displaystyle \mathrm{tg}\, x=t$$

i mamy

$$\begin{array}{rllll} \displaystyle \sin^2x & = & \displaystyle\frac{\mathrm{tg}\,^2x}{1+\mathrm{tg}\,^2x} & = & \displaystyle\frac{t^2}{1+t^2}, \\ \\ \cos^2x & = & \displaystyle\frac{1}{1+\mathrm{tg}\,^2x} & = & \displaystyle\frac{1}{1+t^2}, \\ \\ \sin x\cos x & = & \displaystyle\frac{\mathrm{tg}\, x}{1+\mathrm{tg}\,^2x} & = & \displaystyle\frac{t}{1+t^2} \end{array}$$

oraz

$$\displaystyle x \ =\ \mathrm{arctg}\, t,\quad \,dx=\frac{\,dt}{1+t^2}.$$

Zatem po podstawieniu dostajemy całkę

$$\displaystyle \int R\bigg(\frac{t^2}{1+t^2}, \frac{1}{1+t^2}, \frac{t}{1+t^2} \bigg)\frac{dt}{1+t^2}.$$

Obliczyć całkę $$\displaystyle \int\frac{dx}{1+2\cos^2x}\,dx.$$
W całce tej stosujemy podstawienie $$\displaystyle \mathrm{tg}\, x=t,$$ wówczas $$\displaystyle \cos^2x=\frac{1}{1+t^2}$$ i $$\displaystyle dx=\frac{\,dt}{1+t^2}.$$ Zatem
$$\displaystyle \int\frac{dx}{1+2\cos^2x}\,dx \ =\ \int\frac{\displaystyle\frac{1}{1+t^2}}{\displaystyle 1+\frac{2}{1+t^2}}\,dt \ =\ \int\frac{dt}{t^2+3}.$$
$$\displaystyle \int\frac{dx}{1+2\cos^2x}\,dx \ =\ \mathrm{arctg}\,\bigg(\frac{\mathrm{tg}\, x}{\sqrt{3}}\bigg)+c.$$