Question
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{4}}_{0}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$

Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$

$\int\limits(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}=2\tan\text{x}+\frac{\text{x}^{4}}{4}+2\text{x}=\text{F}\text{(x)}$

By second fundamental theorem of calculus, we obtain

$\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$

$=\left\{\Bigg(2\tan\frac{\pi}{4}+\frac{1}{4}\bigg(\frac{\pi}{4}\bigg)^{4}+2\bigg(\frac{\pi}{4}\bigg)\Bigg)-(2\tan0+0+0)\right\}$

$=2\tan\frac{\pi}{4}+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2}$

$=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}$

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