5.4 Indefinite Integrals and the Net Change Theorem/17: Difference between revisions
No edit summary |
No edit summary |
||
| Line 11: | Line 11: | ||
<math> | <math> | ||
\int(1+\tan^2{\alpha})\,d\alpha = \int\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)d\alpha = \int\frac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}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 | ||
\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\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 \cos^2x+sin^2x=1 \int\frac{1}{cos^2x}dx = \int\sec^2xdx = \tan{x}+C }