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


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


\tan{x}+C
\tan{x}+C

Revision as of 17:49, 13 September 2022


Note:


Or,

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int (1+\tan ^{2}{\alpha })\,d\alpha =\int \left(1+{\frac {sin^{2}\alpha }{cos^{2}\alpha }}\right)d\alpha =\int \left({\frac {cos^{2}\alpha +sin^{2}\alpha }{cos^{2}\alpha }}\right)d\alpha =\int {\frac {1}{cos^{2}\alpha }}d\alpha =\int \sec ^{2}\alpha \,d\alpha =\tan {x}+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 \cos^2\alpha+sin^2\alpha=1}