What is the integral of x/(sin(x^2)) ?

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

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integral of x/(sin(x^2))

\int \frac{x}{sin ( x ^{2} )} d x

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

Solution steps

\int \frac{x}{sin ( x ^{2} )} d x

Rewrite using trig identities

= \int x csc (x^{2}) d x

Apply u-substitution

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

= \int \frac{csc ( 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 csc (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 tan (\frac{x ^{2}}{2})

Add a constant to the solution

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

y = x^{2} - 2 x + 1, (- 1, 4)

\frac{\partial}{\partial x} (5^{8^{x^{2}}})

\int (e^{x} + 1) \cdot (x - 4) d x

f (x) = 25 - x^{2}, (5, 0)

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