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

Revision as of 07:14, 29 August 2022

17) $\int _{}^{}1+tan^{2}xdx$ =

$\int _{}^{}1+{\frac {sin^{2}x}{cos^{2}x}}dx$ =

$\int _{}^{}{\frac {cos^{2}x+sin^{2}x}{cos^{2}x}}dx$

$\cos ^{2}x+sin^{2}x=1$ thus,

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

$\int _{}^{}\sec ^{2}xdx$

$tanx+C$

@@ Line 3: / Line 3: @@
 <math>\int_{}^{}1+\frac{sin^2x}{cos^2x}dx</math> =
-<math>\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx</math> <math>\cos^2x+sin^2x=1</math> thus,
+<math>\int_{}^{}\frac{cos^2x+sin^2x}{cos^2x}dx</math>
+<math>\cos^2x+sin^2x=1</math> thus,
 <math>\int_{}^{}\frac{1}{cos^2x}dx</math> =
+<math>\int_{}^{}\sec^2xdx</math>
+<math>tanx+C</math>
 [[5.3 The Fundamental Theorem of Calculus/1|1]]