What is the integral of (sin(θ))/(3+cos^2(θ)) ?

The integral of (sin(θ))/(3+cos^2(θ)) is -1/(sqrt(3))arctan((cos(θ))/(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 (sin(θ))/(3+cos^2(θ))

\int \frac{sin ( θ )}{3 + cos ^{2} ( θ )} d θ

- \frac{1}{3} arctan (\frac{cos ( θ )}{3}) + C

Solution steps

\int \frac{sin ( θ )}{3 + cos ^{2} ( θ )} d θ

Apply u-substitution

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

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

Take the constant out:

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

= - \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{cos ( θ )}{3})

Add a constant to the solution

= - \frac{1}{3} arctan (\frac{cos ( θ )}{3}) + C

\int \frac{9}{x ^{2} 9 - x ^{2}} d x

\int 2 e^{- x} + \frac{6}{x} + ln (2) d x

\int \frac{- x ^{- 3} + 1}{x ^{3}} d x

\int (sin^{3} (x) cos^{2} (x)) d x

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