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

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<math>
<math>


\int(1+\tan^2{\alpha})\,d\alpha
\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  
 
\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
\cos^2x+sin^2x=1

Revision as of 17:48, 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 }