What is the integral of 2/(3cos^2(x)+1) ?

The integral of 2/(3cos^2(x)+1) is arctan(1/2 tan(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(3cos^2(x)+1)

\int \frac{2}{3 cos ^{2} ( x ) + 1} d x

arctan (\frac{1}{2} tan (x)) + C

Solution steps

\int \frac{2}{3 cos ^{2} ( x ) + 1} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{3 cos ^{2} ( x ) + 1} d x

Apply Trigonometric Substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{tan ( x )}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{tan ( x )}{2}) : arctan (\frac{1}{2} tan (x))

= arctan (\frac{1}{2} tan (x))

Add a constant to the solution

= arctan (\frac{1}{2} tan (x)) + C

\int (x + 3) 7^{(x + 3)^{2}} d x

\int 3 x^{2} y^{2} + \frac{x}{y} d y

\int (tan^{4} (x) - sec^{4} (x)) d x

\int 4 - 2^{- x} d x

\int x^{22} d x

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