What is the integral of (cot(ln(x^4+1))x^3)/(x^4+1) ?

The integral of (cot(ln(x^4+1))x^3)/(x^4+1) is 1/4 ln|sin(ln(x^4+1))|+C

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integral of (cot(ln(x^4+1))x^3)/(x^4+1)

\int \frac{cot ( ln ( x ^{4} + 1 ) ) x ^{3}}{x ^{4} + 1} d x

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

Solution steps

\int \frac{cot ( ln ( x ^{4} + 1 ) ) x ^{3}}{x ^{4} + 1} d x

Apply u-substitution

: \int \frac{cot ( ln ( u ) )}{4 u} d u

= \int \frac{cot ( ln ( u ) )}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cot ( ln ( u ) )}{u} d u

Apply u-substitution

: \int cot (v) d v

= \frac{1}{4} \cdot \int cot (v) d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( v )}{sin ( v )} d v

Apply u-substitution

: \int \frac{1}{w} d w

= \frac{1}{4} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{4} ln ∣ w ∣

Substitute back

= \frac{1}{4} ln sin (ln (x^{4} + 1))

Add a constant to the solution

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

\frac{( s ^{2} + 1 )}{s ( s - 1 ) ^{3}}

\frac{\partial}{\partial x} (e^{x y + 1})

\frac{( 5 x - 2 )}{( x + 1 )}, 4

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

3 cos (x), 3 cos (2 x), 0 \leq x \leq π

What is the integral of (cot(ln(x^4+1))x^3)/(x^4+1) ?
The integral of (cot(ln(x^4+1))x^3)/(x^4+1) is 1/4 ln|sin(ln(x^4+1))|+C