What is the integral of (sec(x)tan(x))/(9+4sec^2(x)) ?

The integral of (sec(x)tan(x))/(9+4sec^2(x)) is 1/6 arctan(2/3 sec(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sec(x)tan(x))/(9+4sec^2(x))

\int \frac{sec ( x ) tan ( x )}{9 + 4 sec ^{2} ( x )} d x

\frac{1}{6} arctan (\frac{2}{3} sec (x)) + C

Solution steps

\int \frac{sec ( x ) tan ( x )}{9 + 4 sec ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{9 + 4 u ^{2}} d u

= \int \frac{1}{9 + 4 u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{2}{3} sec (x))

Add a constant to the solution

= \frac{1}{6} arctan (\frac{2}{3} sec (x)) + C

\int cos (t) e^{t} d t

\frac{d x}{d y} = y^{2} x - x + y^{2} - 1

\int tan^{2} (x) sec^{3} (d) x d x

f (x) = arctan (x)

\int \frac{e ^{2 x}}{e ^{x} + 1} d x

What is the integral of (sec(x)tan(x))/(9+4sec^2(x)) ?
The integral of (sec(x)tan(x))/(9+4sec^2(x)) is 1/6 arctan(2/3 sec(x))+C