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

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

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

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

1

Solution steps

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

Simplify

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

Use the common integral:

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

= [tan (t)]_{0}^{\frac{π}{4}}

Compute the boundaries:

1

= 1

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

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (x) d x

\int_{0}^{x} \frac{t}{t ^{2} + 1} d t

\int_{2}^{8} \frac{1}{x} d x

\int_{0}^{1} x^{6} d x

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