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

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<math>\int1+\frac{sin^2x}{cos^2x}\,dx<math>
<math>\int1+\frac{sin^2x}{cos^2x}\,dx<math>
<math>\int\frac{cos^2x+sin^2x}{cos^2x}\,dx<math>  
<math>\int\frac{cos^2x+sin^2x}{cos^2x}\,dx<math>  
<math>\cos^2x+sin^2x=1<math>, thus
<math>\cos^2x+sin^2x=1<math>
<math>\int\frac{1}{cos^2x}\,dx<math>
<math>\int\frac{1}{cos^2x}\,dx<math>
<math>\int\sec^2x\,dx<math>
<math>\int\sec^2x\,dx<math>
<math>tanx+C\<math>
<math>tanx+C\<math>
\end{align}
\end{align}

Revision as of 03:48, 29 August 2022

17. \begin{align}

<math>\int_{}^{}1+tan^2 x\,dx <math> <math>\int1+\frac{sin^2x}{cos^2x}\,dx<math> <math>\int\frac{cos^2x+sin^2x}{cos^2x}\,dx<math> <math>\cos^2x+sin^2x=1<math> <math>\int\frac{1}{cos^2x}\,dx<math> <math>\int\sec^2x\,dx<math> <math>tanx+C\<math> \end{align}