What is the integral of sin(4x)ln(sin(4x)) ?

The integral of sin(4x)ln(sin(4x)) is 1/4 (-ln(sin(4x))cos(4x)+ln|tan(2x)|+cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(4x)ln(sin(4x))

\int sin (4 x) ln (sin (4 x)) d x

\frac{1}{4} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x)) + C

Solution steps

\int sin (4 x) ln (sin (4 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{4} d u

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

Apply Integration By Parts

: - ln (sin (u)) cos (u) - \int - cot (u) cos (u) d u

= \frac{1}{4} (- ln (sin (u)) cos (u) - \int - cot (u) cos (u) d u)

\int - cot (u) cos (u) d u = - ln tan (\frac{u}{2}) - cos (u)

= \frac{1}{4} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 4 x

= \frac{1}{4} (- ln (sin (4 x)) cos (4 x) - (- ln tan (\frac{4 x}{2}) - cos (4 x)))

Simplify

\frac{1}{4} (- ln (sin (4 x)) cos (4 x) - (- ln tan (\frac{4 x}{2}) - cos (4 x))) : \frac{1}{4} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x))

= \frac{1}{4} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x))

Add a constant to the solution

= \frac{1}{4} (- ln (sin (4 x)) cos (4 x) + ln ∣ tan (2 x) ∣ + cos (4 x)) + C

\int (3 t + t^{2}) sin (2 t) d t

\frac{d}{d x} (6 x x^{3} + 3 x)

\frac{\partial}{\partial x} (x h)

y^{^{'}} + 6 y = t + e^{- 2 t}

\frac{d}{d x} (\frac{5 x}{1 + 2 x ^{2}})

What is the integral of sin(4x)ln(sin(4x)) ?
The integral of sin(4x)ln(sin(4x)) is 1/4 (-ln(sin(4x))cos(4x)+ln|tan(2x)|+cos(4x))+C