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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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{sin ^{2} ( x )}{2})

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x + 2}{3})

x \to 0 lim (\frac{tan ( π x )}{x})

\frac{d}{d x} (e^{y z (x)})

x \to 2 lim ([4 f (- 8) + g (2)])

n = 1 \sum \infty \frac{- 2 + 3 ^{n}}{2 ^{2 n}}

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