What is the integral from 0 to pi/3 of sec(θ)tan(θ) ?

The integral from 0 to pi/3 of sec(θ)tan(θ) is 1

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integral from 0 to pi/3 of sec(θ)tan(θ)

\int_{0}^{\frac{π}{3}} sec (θ) tan (θ) d θ

1

Solution steps

\int_{0}^{\frac{π}{3}} sec (θ) tan (θ) d θ

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:

1

= 1

\int_{0}^{2 π} 1 + x^{2} d x

\int_{1}^{3 x} x^{3} e^{x y} d y

\int_{0}^{1 - y} 2 d x

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

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

What is the integral from 0 to pi/3 of sec(θ)tan(θ) ?
The integral from 0 to pi/3 of sec(θ)tan(θ) is 1