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


\end{align}
\end{align}
</math>
<math>
</math>
</math>

Revision as of 16:03, 21 September 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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] & =left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan\left({0}\right) + 0\right]\\[2ex] & =1+\frac{\pi}{4} \end{align} }