What is the integral from-pi/4 to 0 of 6sec^3(x) ?

The integral from-pi/4 to 0 of 6sec^3(x) is 3(-ln(sqrt(2)-1)+sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-pi/4 to 0 of 6sec^3(x)

\int_{- \frac{π}{4}}^{0} 6 sec^{3} (x) d x

3 (- ln (2 - 1) + 2)

Decimal

6.88676 \dots

Solution steps

\int_{- \frac{π}{4}}^{0} 6 sec^{3} (x) d x

Take the constant out:

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

= 6 \cdot \int_{- \frac{π}{4}}^{0} sec^{3} (x) d x

Apply Integral Reduction

= 6 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{0} sec (x) d x)

\int_{- \frac{π}{4}}^{0} sec (x) d x = - ln (2 - 1)

= 6 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} (- ln (2 - 1)))

Simplify

6 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} (- ln (2 - 1))) : 6 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{4}}^{0} - \frac{1}{2} ln (2 - 1))

= 6 ([\frac{1}{2} sec (x) tan (x)]_{- \frac{π}{4}}^{0} - \frac{1}{2} ln (2 - 1))

Compute the boundaries:

\frac{1}{2}

= 6 (\frac{1}{2} - \frac{1}{2} ln (2 - 1))

Simplify

= 3 (- ln (2 - 1) + 2)

x \to 2 - lim (\frac{∣ x - 2 ∣}{x - 2})

\frac{\partial}{\partial x} (\frac{18 x y}{x + y})

x \to 8 lim (15)

\int sin (x) ln (sin (x)) d x

x^{6}, at x = 3

What is the integral from-pi/4 to 0 of 6sec^3(x) ?
The integral from-pi/4 to 0 of 6sec^3(x) is 3(-ln(sqrt(2)-1)+sqrt(2))