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

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= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]
= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]


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


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

Revision as of 16:06, 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] &= \left\tan({\theta}) + \theta \right]\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} \end{align} }