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

From Burton Tech. Points Wiki
Jump to navigation Jump to search
No edit summary
m (Protected "5.4 Indefinite Integrals and the Net Change Theorem/37" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite)))
 
(13 intermediate revisions by the same user not shown)
Line 4: Line 4:
\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)}{\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(\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({\theta}) + \theta)\Bigg|_{0}^{\frac{\pi}{4}}\\[2ex]
& =\tan({\frac{\pi}{4}}) + \frac{\pi}{4} \\[2ex]
& =1+\frac{\pi}{4}
 


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


&= 1+\frac{\pi}{4}


\end{align}
\end{align}
</math>
<math>
= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta
</math>
</math>

Latest revision as of 19:41, 21 September 2022