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

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

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

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

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

Solution steps

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

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 ( x )}{3})

Add a constant to the solution

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

ln (3 x + 1)

x \to 0 + lim (sin (\frac{1}{x}))

\int_{0}^{4} 6 x + \frac{1}{( x + 3 ) ^{2}} d x

\frac{d}{d x} (- x^{3} + 9 x^{2} - 24 x + 5)

\frac{\partial}{\partial x} (3 x^{2} - 2 x y - 4 y^{2})

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