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



Revision as of 16:02, 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]&=\tan \left({\frac {\pi }{4}}\right)+{\frac {\pi }{4}}\\[2ex]&=1+{\frac {\pi }{4}}\end{aligned}}}

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