What is the integral of xcot(x^2) ?

The integral of xcot(x^2) is 1/2 ln|sin(x^2)|+C

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integral of xcot(x^2)

\int x cot (x^{2}) d x

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

Solution steps

\int x cot (x^{2}) d x

Apply u-substitution

: \int \frac{cot ( u )}{2} d u

= \int \frac{cot ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cot (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{cos ( u )}{sin ( u )} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

y^{^{'}} + y = e^{- t}

n = 1 \sum \infty \frac{n - 5}{n} + 5

\int \frac{x ^{2}}{x ^{2} + 6} d x

\int (e^{- \frac{1}{x ^{2}} - 3 l n (x)} \cdot 1) d x

\int \frac{( 3 x - 1 )}{x ^{2} + 9} d x

What is the integral of xcot(x^2) ?
The integral of xcot(x^2) is 1/2 ln|sin(x^2)|+C