5.4 Indefinite Integrals and the Net Change Theorem/17: Difference between revisions
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<math>\int_{}^{}1+tan^2xdx = | <math> | ||
\int(1+\tan^2{\alpha})\,d\alpha = \int\sec^2\alpha \,d\alpha = \tan\alpha + C | |||
</math> | |||
OR | |||
<math> | |||
\int_{}^{}1+tan^2xdx = | |||
\int_{}^{}1+\frac{sin^2x}{cos^2x}dx = | \int_{}^{}1+\frac{sin^2x}{cos^2x}dx = | ||
Revision as of 17:45, 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 }
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^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 }