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

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17)<math>\int_{}^{}1+tan^2xdx</math> =  
17)<math>\int_{}^{}1+tan^2xdx =  


<math>\int_{}^{}1+\frac{sin^2x}{cos^2x}dx</math> =  
\int_{}^{}1+\frac{sin^2x}{cos^2x}dx =  


<math>\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx</math>
\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx  


<math>\cos^2x+sin^2x=1</math> thus,
\cos^2x+sin^2x=1


<math>\int_{}^{}\frac{1}{cos^2x}dx</math> =  
\int_{}^{}\frac{1}{cos^2x}dx =  


<math>\int_{}^{}\sec^2xdx</math> =
\int_{}^{}\sec^2xdx =


<math>tanx+C</math>
tanx+C
</math>

Revision as of 17:41, 13 September 2022

17)Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{}^{}1+tan^{2}xdx=\int _{}^{}1+{\frac {sin^{2}x}{cos^{2}x}}dx=\int _{}^{}{\frac {cos^{2}x+sin^{2}x}{cos^{2}x}}dx\cos ^{2}x+sin^{2}x=1\int _{}^{}{\frac {1}{cos^{2}x}}dx=\int _{}^{}\sec ^{2}xdx=tanx+C}

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