What is the integral of (sin^3(x))/(1+cos^2(x)) ?

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

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integral of (sin^3(x))/(1+cos^2(x))

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

- 2 arctan (cos (x)) + cos (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply Trigonometric Substitution

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

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

\frac{1 - u ^{2}}{u ^{2} + 1} = \frac{2}{u ^{2} + 1} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

\int 1 d u = u

= - (2 arctan (u) - u)

Substitute back

u = cos (x)

= - (2 arctan (cos (x)) - cos (x))

Simplify

= - 2 arctan (cos (x)) + cos (x)

Add a constant to the solution

= - 2 arctan (cos (x)) + cos (x) + C

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