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I want to solve $\displaystyle \quad \int\frac{dz}{1+5z^2}$ and Wolfram says I need to use the substitution with inverse tan, taking $u=\sqrt{5}z$, giving me $$\frac{1}{\sqrt{5}}\int\frac{du}{1+u^2}.$$ Surely this is incorrect as the 1 in the demoninator would now be $\sqrt{5}$ too. What don't I understand? |
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If $\;u = \sqrt{5}z\;$ then $\;du = \sqrt{5}\, dz \;\implies\; \color{red}{\bf{dz = \dfrac{1}{\sqrt5}} \,du}\;$ and $$\int\frac{1}{1+5z^2}\,dz = \int \frac{1}{1 + (\sqrt{5}z)^2} \,\color{red}{\bf dz}= \int\frac{1}{1+u^2} \cdot \color{red}{\bf{\frac{1}{\sqrt{5}}\,du}}\; = \;\dfrac{1}{\sqrt{5}} \int \dfrac{1}{1+u^2}\,du \;= \;\;?$$ |
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$du=\sqrt 5 dz$, you have to account for this when substituting |
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