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Question Number 140548 by liberty last updated on 09/May/21

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)=\left|\begin{array}{c}\sec {}^2\mathrm{x}11\\ \cos {}^2\mathrm{x}\cos {}^2\mathrm{x}{\csc }^2\mathrm{x}\\ 1\cos {}^2\mathrm{x}\tan {}^2\mathrm{x}\end{array}\right|$$$$\mathrm{evaluate}\underset{0}{\stackrel{\pi /4}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.$$

Answered by EDWIN88 last updated on 09/May/21

$$\mathrm{I}=\underset{0}{\stackrel{\pi /4}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\left|\begin{array}{c}\underset{0}{\stackrel{\pi /4}{\int }}\sec {}^2\mathrm{x}\mathrm{dx}11\\ \underset{0}{\stackrel{\pi /4}{\int }}\cos {}^2\mathrm{x}\mathrm{dx}\underset{0}{\stackrel{\pi /4}{\int }}\cos {}^2\mathrm{x}\mathrm{dx}\underset{0}{\stackrel{\pi /4}{\int }}{\csc }^2\mathrm{x}\mathrm{dx}\\ 1\underset{0}{\stackrel{\pi /4}{\int }}\cos {}^2\mathrm{x}\mathrm{dx}\underset{0}{\stackrel{\pi /4}{\int }}{\tan }^2\mathrm{x}\mathrm{dx}\end{array}\right|$$$$\left(1\right)\underset{0}{\stackrel{\pi /4}{\int }}\sec {}^2\mathrm{x}\mathrm{dx}=1\left(2\right)\underset{0}{\stackrel{\pi /4}{\int }}\cos {}^2\mathrm{x}\mathrm{dx}=\underset{0}{\stackrel{\pi /4}{\int }}\left(\frac{1+\cos 2\mathrm{x}}{2}\right)\mathrm{dx}={\left[\frac{\mathrm{x}+\frac{\sin 2\mathrm{x}}{2}}{2}\right]}_0^{\pi /4}=\frac{\pi +2}{8}$$$$\left(3\right)\underset{0}{\stackrel{\pi /4}{\int }}{\csc }^2\mathrm{x}\mathrm{dx}=-1\left(4\right)\underset{0}{\stackrel{\pi /4}{\int }}{\tan }^2\mathrm{x}\mathrm{dx}=\underset{0}{\stackrel{\pi /4}{\int }}(\sec {}^2\mathrm{x}-1)\mathrm{dx}={[\tan \mathrm{x}-\mathrm{x}]}_0^{\pi /4}=\frac{4-\pi }{4}$$$$\mathrm{I}=\left|\begin{array}{c}111\\ \frac{\pi +2}{8}\frac{\pi +2}{8}-1\\ 1\frac{\pi +2}{8}\frac{4-\pi }{4}\end{array}\right|$$

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