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

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<math>
= \int_{0}^{\frac{\pi}{4}}\sec^2(\theta)} + 1 \\[2ex]
= \int_{0}^{\frac{\pi}{4}}\sec^2(\theta)} + 1  
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</math>

Revision as of 16:01, 21 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 \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 & =\tan({\theta}) + \theta \ \bigg|_{0}^{\frac{\pi}{4}}\\[2ex] & =\tan({\frac{\pi}{4}}) + \frac{\pi}{4} \\[2ex] & =1+\frac{\pi}{4} \end{align} }

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_{0}^{\frac{\pi}{4}}\sec^2(\theta)} + 1 }