What is the integral of 1/(4-cos^2(x)) ?

The integral of 1/(4-cos^2(x)) is 1/(2sqrt(3))arctan(2/(sqrt(3))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 1/(4-cos^2(x))

\int \frac{1}{4 - cos ^{2} ( x )} d x

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

Solution steps

\int \frac{1}{4 - cos ^{2} ( x )} d x

Rewrite using trig identities

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

Apply Trigonometric Substitution

: \int \frac{1}{4 u ^{2} + 3} d u

= \int \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{1}{2 3 ( 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}{2 3} \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}{2 3} arctan (v)

Substitute back

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

Add a constant to the solution

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

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

\int 10 cos (t) + sec^{2} (t) d t

\int 4 x^{3} e^{x^{2}} d x

\int d (x^{2}) d x

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

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