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

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17)<math>\int{}{}1+tan^{2}x*dx = \frac{d}{dx}(tan(x))</math>
17. \begin{align}


<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>
&= \int_{}^{}1+tan^2 x\,dx \\[2ex]
 
&= \int1+\frac{sin^2x}{cos^2x}\,dx \\[2ex]
 
&= \int\frac{cos^2x+sin^2x}{cos^2x}\,dx \\[2ex]
<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>
&= \cos^2x+sin^2x=1, thus
 
&= \int\frac{1}{cos^2x}\,dx \\[2ex]
 
&= \int\sec^2x\,dx \\[2ex]
<math>=1+tan^2(x)</math>
&= tanx+C\
 
\end{align}
 
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Revision as of 03:44, 29 August 2022

17. \begin{align}

&= \int_{}^{}1+tan^2 x\,dx \\[2ex] &= \int1+\frac{sin^2x}{cos^2x}\,dx \\[2ex] &= \int\frac{cos^2x+sin^2x}{cos^2x}\,dx \\[2ex] &= \cos^2x+sin^2x=1, thus &= \int\frac{1}{cos^2x}\,dx \\[2ex] &= \int\sec^2x\,dx \\[2ex] &= tanx+C\ \end{align}

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