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

The integral of (tan(x))/(sqrt(4+9tan^2(x))) is -1/(sqrt(5))arcsin((sqrt(5))/3 cos(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{5} arcsin (\frac{5}{3} cos (x)) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{- 5 u ^{2} + 9} d u

= \int - \frac{1}{- 5 u ^{2} + 9} d u

Take the constant out:

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

= - \int \frac{1}{- 5 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{5} d v

= - \int \frac{1}{5} d v

Take the constant out:

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

= - \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= - \frac{1}{5} \cdot 1 \cdot v

Substitute back

= - \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{3} cos (x))

Simplify

= - \frac{1}{5} arcsin (\frac{5}{3} cos (x))

Add a constant to the solution

= - \frac{1}{5} arcsin (\frac{5}{3} cos (x)) + C

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

\int \frac{x}{3} e^{- \frac{x}{3}} d x

\int \frac{3 x ^{2} - 2 x + 19}{( x - 1 ) ( x ^{2} + 9 )} d x

\int sin^{2} (8 n - 3) d n

\int (5 e^{x} + \frac{10}{x} - 4 x^{3} + 2^{x}) d x

What is the integral of (tan(x))/(sqrt(4+9tan^2(x))) ?
The integral of (tan(x))/(sqrt(4+9tan^2(x))) is -1/(sqrt(5))arcsin((sqrt(5))/3 cos(x))+C