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

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

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

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

2 - 1

Decimal

0.41421 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} 1 d u

= \int_{1}^{2} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{1}^{2}

Simplify

= [u]_{1}^{2}

Compute the boundaries:

2 - 1

= 2 - 1

\int_{- 2}^{0} 2 x x^{2} + 9 d x

\int_{- 2}^{0} (2 + x^{2}) d x

\int_{0}^{h} y^{2} d y

\int_{- 1}^{1} \frac{1}{3 x - 4} d x

\int_{e}^{x} sin (t) d t

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