5.4 Indefinite Integrals and the Net Change Theorem/37: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{0}^{\frac{\pi}{4}}\left(\frac{1+\cos^2(\theta)}{\cos^2(\theta)}\right)d\theta \ = \ \int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta)} + \frac{\cos^2(\theta)}{\cos^2(\theta)} \ = \ \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2(\theta)} + 1 | \int_{0}^{\frac{\pi}{4}}\left(\frac{1+\cos^2(\theta)}{\cos^2(\theta)}\right)d\theta \ = \ \int_{0}^{\frac{\pi}{4}}\frac{1}{\cos^2(\theta)} + \frac{\cos^2(\theta)}{\cos^2(\theta)} \ = \ \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2(\theta)} + 1 \\ | ||
=\tan({\theta}) + \theta \ \bigg|_{0}^{\frac{\pi}{4}} | =\tan({\theta}) + \theta \ \bigg|_{0}^{\frac{\pi}{4}}\\ | ||
=\tan({\frac{\pi}{4}}) + \frac{\pi}{4} - \left(\tan(\theta)+0\right) | =\tan({\frac{\pi}{4}}) + \frac{\pi}{4} - \left(\tan(\theta)+0\right)\\ | ||
=1+\frac{\pi}{4} | =1+\frac{\pi}{4} | ||
Revision as of 06:06, 3 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}}{\frac {1}{\cos ^{2}(\theta )}}+{\frac {\cos ^{2}(\theta )}{\cos ^{2}(\theta )}}\ =\ \int _{0}^{\frac {\pi }{4}}{\frac {1}{\cos ^{2}(\theta )}}+1\\=\tan({\theta })+\theta \ {\bigg |}_{0}^{\frac {\pi }{4}}\\=\tan({\frac {\pi }{4}})+{\frac {\pi }{4}}-\left(\tan(\theta )+0\right)\\=1+{\frac {\pi }{4}}\end{aligned}}}