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_{0}^{1} \frac{1}{1 + e ^{x}} d x

\int_{e}^{e^{2}} \frac{1}{x ( ln ( x ) ) ^{2}} d x

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

\int_{0}^{10} e^{x} d x

\int_{2}^{4} t^{2} - 2 t - 3 d t

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