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

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17. \begin{align}
<math>
\int(1+\tan^2{\alpha})\,d\alpha = \int\sec^2\alpha \,d\alpha = \tan\alpha + C
</math>


<math>\int_{}^{}1+tan^2 x\,dx <math>
 
<math>\int1+\frac{sin^2x}{cos^2x}\,dx<math>
Note: <math>1+\tan^2{\alpha} = \sec^2\alpha</math>
<math>\int\frac{cos^2x+sin^2x}{cos^2x}\,dx<math>
 
<math>\cos^2x+sin^2x=1<math>, thus
 
<math>\int\frac{1}{cos^2x}\,dx<math>
Or,
<math>\int\sec^2x\,dx<math>
 
<math>tanx+C\<math>
 
\end{align}
<math>
 
\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\left(\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\right)d\alpha  = \int\frac{1}{cos^2\alpha}\,d\alpha =
 
\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C
</math>
 
 
Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

Latest revision as of 19:39, 21 September 2022


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Or,



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