5.4 Indefinite Integrals and the Net Change Theorem/17: Difference between revisions

Revision as of 00:30, 29 August 2022

17) $\int {}{}1+tan^{2}x*dx={\frac {d}{dx}}(tan(x))$

${\frac {d}{dx}}(tan(x))={\frac {d}{dx}}\left({\frac {sin(x)}{cos(x)}}\right)={\frac {cos^{2}x-(-(sin(x))(sin(x))}{cos^{2}(x)}}$

$={\frac {cos^{2}x+sin^{2}(x))}{cos^{2}(x)}}={\frac {cos^{2}x}{cos^{2}(x)}}+{\frac {sin^{2}x}{cos^{2}(x)}}$

$=1+tan^{2}(x)$

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 )<math>\int{}{}1+tan^{2}x*dx = \frac{d}{dx}(tan(x))</math>
-<math>\frac{d}{dx}(tan(x))=\frac{d}{dx}\left(\frac{sin(x)}{cos(x)}\right)=\frac{cos^{2}x-(-sin(x))}{cos^{2}(x)}</math>
+<math>\frac{d}{dx}(tan(x))=\frac{d}{dx}\left(\frac{sin(x)}{cos(x)}\right)=\frac{cos^{2}x-(-(sin(x))(sin(x))}{cos^{2}(x)}</math>
+<math>=\frac{cos^{2}x+sin^2(x))}{cos^{2}(x)}=\frac{cos^{2}x}{cos^{2}(x)}+\frac{sin^{2}x}{cos^{2}(x)}</math>
+<math>=1+tan^2(x)</math>