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

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<math>
<math>


\int_{}^{}1+tan^2xdx =
\int(1+\tan^2{\alpha})\,d\alpha


\int_{}^{}1+\frac{sin^2x}{cos^2x}dx =  
\int 1+\frac{sin^2\alpha}{cos^2\alpha}d\alpha =  


\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx
\int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha


\cos^2x+sin^2x=1
\cos^2x+sin^2x=1


\int_{}^{}\frac{1}{cos^2x}dx =  
\int\frac{1}{cos^2x}dx =  


\int_{}^{}\sec^2xdx =
\int\sec^2xdx =


tanx+C
\tan{x}+C
</math>
</math>

Revision as of 17:47, 13 September 2022

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^2{\alpha})\,d\alpha = \int\sec^2\alpha \,d\alpha = \tan\alpha + C }


Note: 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 1+\tan^2{\alpha} = \sec^2\alpha}


Or,

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^2{\alpha})\,d\alpha \int 1+\frac{sin^2\alpha}{cos^2\alpha}d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}d\alpha \cos^2x+sin^2x=1 \int\frac{1}{cos^2x}dx = \int\sec^2xdx = \tan{x}+C }