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

The integral of 1/(4sin^2(x)+cos^2(x)) is -1/2 arctan((cot(x))/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4sin^2(x)+cos^2(x))

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

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

Solution steps

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

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= - \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

= - \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)

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

Substitute back

= - \frac{1}{2} arctan (\frac{cot ( x )}{2})

Add a constant to the solution

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

\int \frac{3 x + 23}{x ^{2} - 8 x - 9} d x

\int e^{8 x - 6} d x

\int \frac{( 4 + ln ( x ) ) ^{5}}{4 x} d x

\int (x - π) sin (n x) d x

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