What is the integral from 0 to pi/6 of sqrt(1+(tan(x))^2) ?

The integral from 0 to pi/6 of sqrt(1+(tan(x))^2) is 1/2 ln(3)

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integral from 0 to pi/6 of sqrt(1+(tan(x))^2)

\int_{0}^{\frac{π}{6}} 1 + (tan (x))^{2} d x

\frac{1}{2} ln (3)

Decimal

0.54930 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 1 + (tan (x))^{2} d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} sec^{2} (x) d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{\frac{π}{6}} ∣ sec (x) ∣ d x

Eliminate Absolutes

= \int_{0}^{\frac{π}{6}} sec (x) d x

Use the common integral:

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

= [ln ∣ tan (x) + sec (x) ∣]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{1}{2} ln (3)

= \frac{1}{2} ln (3)

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What is the integral from 0 to pi/6 of sqrt(1+(tan(x))^2) ?
The integral from 0 to pi/6 of sqrt(1+(tan(x))^2) is 1/2 ln(3)