What is the integral of sec^2(3t) ?

The integral of sec^2(3t) is 1/3 tan(3t)+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(3t)

\int sec^{2} (3 t) d t

\frac{1}{3} tan (3 t) + C

Solution steps

\int sec^{2} (3 t) d t

Apply u-substitution

: \int sec^{2} (u) \frac{1}{3} d u

= \int sec^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{3} tan (u)

Substitute back

u = 3 t

= \frac{1}{3} tan (3 t)

Add a constant to the solution

= \frac{1}{3} tan (3 t) + C

\int 9 (3 x - 1)^{2} cos (3 x) d x

\int \frac{x}{( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

\int x + x^{12} + x^{- 12} d x

\int x^{5} \cdot sin (x^{6}) d x

\int \frac{1}{10 ( 1 + x ^{2} )} d x