What is the integral from 0 to pi/6 of (sin(t))/(cos^2(t)) ?

The integral from 0 to pi/6 of (sin(t))/(cos^2(t)) is -1+2/(sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to pi/6 of (sin(t))/(cos^2(t))

\int_{0}^{\frac{π}{6}} \frac{sin ( t )}{cos ^{2} ( t )} d t

- 1 + \frac{2}{3}

Decimal

0.15470 \dots

Solution steps

\int_{0}^{\frac{π}{6}} \frac{sin ( t )}{cos ^{2} ( t )} d t

Apply u-substitution

= \int_{1}^{\frac{3}{2}} - \frac{1}{u ^{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{3}{2}}^{1} - \frac{1}{u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{3}{2}}^{1} \frac{1}{u ^{2}} d u)

Apply the Power Rule

= - (- [- \frac{1}{u}]_{\frac{3}{2}}^{1})

Simplify

= [- \frac{1}{u}]_{\frac{3}{2}}^{1}

Compute the boundaries:

- 1 + \frac{2}{3}

= - 1 + \frac{2}{3}

What is the integral from 0 to pi/6 of (sin(t))/(cos^2(t)) ?
The integral from 0 to pi/6 of (sin(t))/(cos^2(t)) is -1+2/(sqrt(3))