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

Latest revision as of 19:39, 21 September 2022

$\int (1+\tan ^{2}{\alpha })\,d\alpha =\int \sec ^{2}\alpha \,d\alpha =\tan \alpha +C$

Note: $1+\tan ^{2}{\alpha }=\sec ^{2}\alpha$

Or,

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

Note: $\cos ^{2}\alpha +sin^{2}\alpha =1$

@@ Line 1: / Line 1: @@
-. \begin{align}
+<math>
+\int(1+\tan^2{\alpha})\,d\alpha = \int\sec^2\alpha \,d\alpha = \tan\alpha + C
+</math>
-&= \int_{}^{}1+tan^2 x\,dx \\[2ex]
-&= \int1+\frac{sin^2x}{cos^2x}\,dx \\[2ex]
+Note: <math>1+\tan^2{\alpha} = \sec^2\alpha</math>
-&= \int\frac{cos^2x+sin^2x}{cos^2x}\,dx \\[2ex]
-&= \cos^2x+sin^2x=1, thus
-&= \int\frac{1}{cos^2x}\,dx \\[2ex]
+Or,
-&= \int\sec^2x\,dx \\[2ex]
-&= tanx+C\
-\end{align}
+<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\frac{1}{cos^2\alpha}\,d\alpha =
+\int\sec^2\alpha \,d\alpha = \tan{\alpha}+C
+</math>
+Note: <math>\cos^2\alpha+sin^2\alpha=1</math>

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

Latest revision as of 19:39, 21 September 2022

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