What is the integral of sin(7x)ln(sin(7x)) ?

The integral of sin(7x)ln(sin(7x)) is 1/7 (-ln(sin(7x))cos(7x)+ln|tan((7x)/2)|+cos(7x))+C

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integral of sin(7x)ln(sin(7x))

\int sin (7 x) ln (sin (7 x)) d x

\frac{1}{7} (- ln (sin (7 x)) cos (7 x) + ln tan (\frac{7 x}{2}) + cos (7 x)) + C

Solution steps

\int sin (7 x) ln (sin (7 x)) d x

Apply u-substitution

: \int sin (u) ln (sin (u)) \frac{1}{7} d u

= \int sin (u) ln (sin (u)) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \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}{7} (- 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}{7} (- ln (sin (u)) cos (u) - (- ln tan (\frac{u}{2}) - cos (u)))

Substitute back

u = 7 x

= \frac{1}{7} (- ln (sin (7 x)) cos (7 x) - (- ln tan (\frac{7 x}{2}) - cos (7 x)))

Simplify

= \frac{1}{7} (- ln (sin (7 x)) cos (7 x) + ln tan (\frac{7 x}{2}) + cos (7 x))

Add a constant to the solution

= \frac{1}{7} (- ln (sin (7 x)) cos (7 x) + ln tan (\frac{7 x}{2}) + cos (7 x)) + C

y^{^{'}} = 2 (1 + y^{2}) x

\frac{d}{d x} (2 x - 1 - \frac{1}{x})

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

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

\int \frac{4 x}{2 x ^{2} + 3} d x

What is the integral of sin(7x)ln(sin(7x)) ?
The integral of sin(7x)ln(sin(7x)) is 1/7 (-ln(sin(7x))cos(7x)+ln|tan((7x)/2)|+cos(7x))+C