What is the integral of (sec^2(x))/(3+tan^2(x)) ?

The integral of (sec^2(x))/(3+tan^2(x)) is 1/(sqrt(3))arctan((tan(x))/(sqrt(3)))+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(x))/(3+tan^2(x))

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

\frac{1}{3} arctan (\frac{tan ( x )}{3}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 + u ^{2}} d u

= \int \frac{1}{3 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{3 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{tan ( x )}{3})

Add a constant to the solution

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

y^{^{''}} - 4 y^{^{'}} + 3 y = t, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

\frac{d}{d x} (\frac{x ^{2}}{ln ( x )})

\frac{d}{d x} (x^{\frac{3}{5}} x^{3} + 2 x - 6)

\frac{d}{d x} (2 x^{2} - 9 x + 10)

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