What is the integral of (sin(x)cos(x))/(4+cos^4(x)) ?

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

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integral of (sin(x)cos(x))/(4+cos^4(x))

\int \frac{sin ( x ) cos ( x )}{4 + cos ^{4} ( x )} d x

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

Solution steps

\int \frac{sin ( x ) cos ( x )}{4 + cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{u}{4 + u ^{4}} d u

= \int - \frac{u}{4 + u ^{4}} d u

Take the constant out:

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

= - \int \frac{u}{4 + u ^{4}} d u

Apply u-substitution

: \int \frac{1}{2 ( 4 + v ^{2} )} d v

= - \int \frac{1}{2 ( 4 + v ^{2} )} d v

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{2} \cdot \frac{1}{2} arctan (w)

Substitute back

= - \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{cos ^{2} ( x )}{2})

Simplify

- \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{cos ^{2} ( x )}{2}) : - \frac{1}{4} arctan (\frac{1}{2} cos^{2} (x))

= - \frac{1}{4} arctan (\frac{1}{2} cos^{2} (x))

Add a constant to the solution

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

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