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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{}^{}1+tan^2xdx = \int_{}^{}1+\frac{sin^2x}{cos^2x}dx = \int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx \cos^2x+sin^2x=1 \int_{}^{}\frac{1}{cos^2x}dx = \int_{}^{}\sec^2xdx = tanx+C }

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