What is the integral of (6sin(2x))/(1+cos^4(x)) ?

The integral of (6sin(2x))/(1+cos^4(x)) is -6arctan(cos^2(x))+C

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integral of (6sin(2x))/(1+cos^4(x))

\int \frac{6 sin ( 2 x )}{1 + cos ^{4} ( x )} d x

- 6 arctan (cos^{2} (x)) + C

Solution steps

\int \frac{6 sin ( 2 x )}{1 + cos ^{4} ( x )} d x

Take the constant out:

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

= 6 \cdot \int \frac{sin ( 2 x )}{1 + cos ^{4} ( x )} d x

Rewrite using trig identities

= 6 \cdot \int \frac{2 cos ( x ) sin ( x )}{1 + cos ^{4} ( x )} d x

Take the constant out:

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

= 6 \cdot 2 \cdot \int \frac{cos ( x ) sin ( x )}{1 + cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{u}{1 + u ^{4}} d u

= 6 \cdot 2 \cdot \int - \frac{u}{1 + u ^{4}} d u

Take the constant out:

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

= 6 \cdot 2 (- \int \frac{u}{1 + u ^{4}} d u)

Apply u-substitution

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

= 6 \cdot 2 (- \int \frac{1}{2 ( 1 + v ^{2} )} d v)

Take the constant out:

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

= 6 \cdot 2 (- \frac{1}{2} \cdot \int \frac{1}{1 + v ^{2}} d v)

Use the common integral:

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

= 6 \cdot 2 (- \frac{1}{2} arctan (v))

Substitute back

= 6 \cdot 2 (- \frac{1}{2} arctan (cos^{2} (x)))

Simplify

6 \cdot 2 (- \frac{1}{2} arctan (cos^{2} (x))) : - 6 arctan (cos^{2} (x))

= - 6 arctan (cos^{2} (x))

Add a constant to the solution

= - 6 arctan (cos^{2} (x)) + C

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What is the integral of (6sin(2x))/(1+cos^4(x)) ?
The integral of (6sin(2x))/(1+cos^4(x)) is -6arctan(cos^2(x))+C