===דוגמאות פרטית===
# {{left|<math>\begin{align}\int&=\int\left(\sin^2(x)\cos^3(x)-\sin(x)\cos^5(x)\right)\mathrm dx\\&=\int\left(\sin^2(x)\cos^2(x)-\sin(x)\cos^4(x)\right)\cos(x)\mathrm dx\\&=\int\left(\sin^2(x)\left(1-\sin^2(x)\right)-\sin(x)\left(1-\sin^2(x)\right)^2\right)\cos(x)\mathrm dx\end{align}</math>}} נציב <math>y=\sin(x)\implies\mathrm dy=\cos(x)\mathrm dx</math> ואז {{left|<math>\begin{align}\int&=\int\left(y^2\left(1-y^2\right)-y\left(1-y^2\right)^2\right)\mathrm dy\\&=\int\left(y^2-y^4-\frac{2y}2\left(1-y^2\right)^2\right)\mathrm dy\\&=\frac{y^3}3-\frac{y^5}5+\frac12\frac{\left(1-y^2\right)^3}3+c\\&=\frac{\sin^3(x)}3-\frac{\sin^5(x)}5+\frac{\cos^6(x)}6+c\end{align}</math>}}{{משל}}
# {{left|<math>\begin{align}\int&=\int\frac{3\cos(x)\sin^3(x)}{2\cos(x)+\sin^2(x)+3}\mathrm dx\\&=\int\frac{3\cos(x)\sin^2(x)}{2\cos(x)+\sin^2(x)+3}\sin(x)\mathrm dx\\&=\int\frac{3\cos(x)\left(1-\cos^2(x)\right)}{2\cos(x)+\left(1-\cos^2(x)\right)+3}\sin(x)\mathrm dx\end{align}</math>}} נציב <math>y=\cos(x)\implies\mathrm dy=-\sin(x)\mathrm dx</math>. לכן: {{left|<math>\begin{align}\int&=\int\frac{3y\left(1-y^2\right)}{2y+4-y^2}(-\mathrm dy)\\&=-\int\frac{3y^3-3y}{y^2-2y-4}\\&=\int\left(3y+6+\frac{21y+24}{y^2-2y-4}\right)\mathrm dy\\&=\frac32y^2+6y+\int\frac{A\mathrm dy}{y-\frac{2+\sqrt{4+16}}2}+\int\frac{B\mathrm dy}{y-\frac{2-\sqrt{4+16}}2}+c\\&=\frac32y^2+6y+\frac{21+9\sqrt5}2\ln\left|y-1-\sqrt5\right|+\frac{21-9\sqrt5}2\ln\left|y-1+\sqrt5\right|+c\\&=\dots\end{align}</math>}}{{משל}}
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===דוגמה===
חשב חשבו <math>\int\frac{\mathrm dx}{5-3\cos(x)}</math>.
====פתרון====
נציב <math>t=\tan\left(\frac x2\right)</math> ולכן {{left|<math>\begin{align}\int&=\int\frac1{5-3\frac{1-t^2}{1+t^2}}\cdot\frac{2\mathrm dt}{1+t^2}\\&=\int\frac{2\mathrm dt}{5+5t^2-3\left(1-t^2\right)}\\&=\int\frac{\mathrm dt}{1+4t^2}\\&=\frac12\arctan(2t)+c\\&=\frac12\arctan\left(2\tan\left(\frac x2\right)\right)+c\end{align}</math>}}
{{משל}}