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

The integral of (xln(sin(x^2)))/(tan(x^2)) is 1/4 ln^2(sin(x^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{ln ( sin ( u ) )}{2 tan ( u )} 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 \frac{ln ( sin ( u ) )}{tan ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{ln ( v )}{v} d v

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{2} \cdot \frac{w ^{2}}{2}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of (xln(sin(x^2)))/(tan(x^2)) ?
The integral of (xln(sin(x^2)))/(tan(x^2)) is 1/4 ln^2(sin(x^2))+C