What is the integral of (5-3sin(2x))/(sec(6x)) ?

The integral of (5-3sin(2x))/(sec(6x)) is 5/6 sin(6x)-3/2 (-1/8 cos(8x)+1/4 cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5-3sin(2x))/(sec(6x))

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

\frac{5}{6} sin (6 x) - \frac{3}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x)) + C

Solution steps

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

Expand

\frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} : \frac{5}{sec ( 6 x )} - \frac{3 sin ( 2 x )}{sec ( 6 x )}

= \int \frac{5}{sec ( 6 x )} - \frac{3 sin ( 2 x )}{sec ( 6 x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5}{sec ( 6 x )} d x - \int \frac{3 sin ( 2 x )}{sec ( 6 x )} d x

\int \frac{5}{sec ( 6 x )} d x = \frac{5}{6} sin (6 x)

\int \frac{3 sin ( 2 x )}{sec ( 6 x )} d x = \frac{3}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x))

= \frac{5}{6} sin (6 x) - \frac{3}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{5}{6} sin (6 x) - \frac{3}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{4} cos (4 x)) + C

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

\frac{1}{4 + x ^{2}}

\int_{0}^{\frac{π}{2}} (sin^{3} (x)) d x

n = 2 \sum \infty \frac{n}{3 ^{n + 1}}

ln (\frac{8 - x}{8 + x})

What is the integral of (5-3sin(2x))/(sec(6x)) ?
The integral of (5-3sin(2x))/(sec(6x)) is 5/6 sin(6x)-3/2 (-1/8 cos(8x)+1/4 cos(4x))+C