What is area (sec^2(x))/2 ,2cos^2(x) ?

The answer to area (sec^2(x))/2 ,2cos^2(x) is diverges

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area (sec^2(x))/2 ,2cos^2(x)

area

\frac{sec ^{2} ( x )}{2}, 2 cos^{2} (x)

diverges

Solution steps

Apply the area formula:

\int_{- \frac{5 π}{4}}^{\frac{5 π}{4}} \frac{sec ^{2} ( x )}{2} - 2 cos^{2} (x) d x

Solve

\int_{- \frac{5 π}{4}}^{\frac{5 π}{4}} \frac{sec ^{2} ( x )}{2} - 2 cos^{2} (x) d x :

diverges

The area is:

= diverges

x \to 2 lim (4 x^{2} - 11 x - 2)

\int_{0}^{1} \frac{1}{( 16 + x ^{2} ) ^{2}} d x

x \to 0 lim (\frac{1}{5 + 5})

\int \frac{1}{1 - ( \frac{x}{L} )} d x

\frac{d y}{d x} + 2 y = 1

What is area (sec^2(x))/2 ,2cos^2(x) ?
The answer to area (sec^2(x))/2 ,2cos^2(x) is diverges