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

The integral of (sin^3(y))/(1+cos(y)) is 4(-cos^2(y/2)+1/2 cos^4(y/2))+C

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integral of (sin^3(y))/(1+cos(y))

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

4 (- cos^{2} (\frac{y}{2}) + \frac{1}{2} cos^{4} (\frac{y}{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{8 u ^{3}}{( 1 + u ^{2} ) ^{3}} d u

= \int \frac{8 u ^{3}}{( 1 + u ^{2} ) ^{3}} d u

Take the constant out:

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

= 8 \cdot \int \frac{u ^{3}}{( 1 + u ^{2} ) ^{3}} d u

Apply u-substitution

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

= 8 \cdot \int \frac{v - 1}{2 v ^{3}} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{v - 1}{v ^{3}} d v

Expand

\frac{v - 1}{v ^{3}} : \frac{1}{v ^{2}} - \frac{1}{v ^{3}}

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2}} - \frac{1}{v ^{3}} d v

Apply the Sum Rule:

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

= 8 \cdot \frac{1}{2} (\int \frac{1}{v ^{2}} d v - \int \frac{1}{v ^{3}} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

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

= 8 \cdot \frac{1}{2} (- \frac{1}{v} - (- \frac{1}{2 v ^{2}}))

Substitute back

= 8 \cdot \frac{1}{2} (- \frac{1}{1 + tan ^{2} ( \frac{y}{2} )} - (- \frac{1}{2 ( 1 + tan ^{2} ( \frac{y}{2} ) ) ^{2}}))

Simplify

8 \cdot \frac{1}{2} (- \frac{1}{1 + tan ^{2} ( \frac{y}{2} )} - (- \frac{1}{2 ( 1 + tan ^{2} ( \frac{y}{2} ) ) ^{2}})) : 4 (- cos^{2} (\frac{y}{2}) + \frac{1}{2} cos^{4} (\frac{y}{2}))

= 4 (- cos^{2} (\frac{y}{2}) + \frac{1}{2} cos^{4} (\frac{y}{2}))

Add a constant to the solution

= 4 (- cos^{2} (\frac{y}{2}) + \frac{1}{2} cos^{4} (\frac{y}{2})) + C

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What is the integral of (sin^3(y))/(1+cos(y)) ?
The integral of (sin^3(y))/(1+cos(y)) is 4(-cos^2(y/2)+1/2 cos^4(y/2))+C