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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-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 (tan^{2} (\frac{x}{2}))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{3} - 7 x^{2})

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

\frac{d}{d x} (\frac{x + 2}{x ^{2} + 4 x + 13})

\frac{\partial}{\partial y} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}})

d er i v a t i v e f (x) = sin^{2} (tan (x))

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