What is the integral of sec^2(8θ) ?

The integral of sec^2(8θ) is 1/8 tan(8θ)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^2(8θ)

\int sec^{2} (8 θ) d θ

\frac{1}{8} tan (8 θ) + C

Solution steps

\int sec^{2} (8 θ) d θ

Apply u-substitution

: \int sec^{2} (u) \frac{1}{8} d u

= \int sec^{2} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{8} tan (u)

Substitute back

u = 8 θ

= \frac{1}{8} tan (8 θ)

Add a constant to the solution

= \frac{1}{8} tan (8 θ) + C

\int x^{3} ln (2 x) d x

\int (6 x + 2)^{2} d x

\int \frac{( cos ( x ) - x sin ( x ) )}{x cos ( x )} d x

\int ∣ 3 x ∣ d x

\int (\frac{3}{1 - x ^{2}}) d x