What is the integral from 0 to pi/4 of (sec(x))^2 ?

The integral from 0 to pi/4 of (sec(x))^2 is 1

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integral from 0 to pi/4 of (sec(x))^2

\int_{0}^{\frac{π}{4}} (sec (x))^{2} d x

1

Solution steps

\int_{0}^{\frac{π}{4}} (sec (x))^{2} d x

Simplify

= \int_{0}^{\frac{π}{4}} sec^{2} (x) d x

Use the common integral:

\int sec^{2} (x) d x = tan (x)

= [tan (x)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= 1

\int_{- π}^{0} cos (n x) d x

\int_{0}^{2} x sin (\frac{nπ x}{2}) d x

\int_{1}^{5} \frac{1}{2 x - 1} d x

\int_{0}^{2 π} cos (θ) sin (θ) d θ

\int_{1}^{e} \frac{1}{x ln ( x )} d x

What is the integral from 0 to pi/4 of (sec(x))^2 ?
The integral from 0 to pi/4 of (sec(x))^2 is 1