5.4 Indefinite Integrals and the Net Change Theorem/37: Difference between revisions

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&= \tan({\theta}) + \theta \ \bigg|_{0}^{\frac{\pi}{4}}\\[2ex]
&= \tan({\theta}) + \theta \ \bigg|_{0}^{\frac{\pi}{4}}\\[2ex]


&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]
&= 1+\frac{\pi}{4}
&= 1+\frac{\pi}{4}


\end{align}
\end{align}
</math>
</math>
&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]

Revision as of 16:04, 21 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{4}}\left({\frac {1+\cos ^{2}(\theta )}{\cos ^{2}(\theta )}}\right)d\theta &=\int _{0}^{\frac {\pi }{4}}\left({\frac {1}{\cos ^{2}(\theta )}}+{\frac {\cos ^{2}(\theta )}{\cos ^{2}(\theta )}}\right)d\theta =\int _{0}^{\frac {\pi }{4}}\left(\sec ^{2}(\theta )+1\right)d\theta \\[2ex]&=\tan({\theta })+\theta \ {\bigg |}_{0}^{\frac {\pi }{4}}\\[2ex]&=1+{\frac {\pi }{4}}\end{aligned}}}

&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]