What is the integral of-9cot(x)ln(sin(x)) ?

The integral of-9cot(x)ln(sin(x)) is -9/2 ln^2(sin(x))+C

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integral of-9cot(x)ln(sin(x))

\int - 9 cot (x) ln (sin (x)) d x

- \frac{9}{2} ln^{2} (sin (x)) + C

Solution steps

\int - 9 cot (x) ln (sin (x)) d x

Take the constant out:

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

= - 9 \cdot \int cot (x) ln (sin (x)) d x

Rewrite using trig identities

= - 9 \cdot \int cos (x) ln (sin (x)) csc (x) d x

Apply u-substitution

: \int \frac{ln ( u )}{u} d u

= - 9 \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= - 9 \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= - 9 \cdot \frac{v ^{2}}{2}

Substitute back

= - 9 \cdot \frac{ln ^{2} ( sin ( x ) )}{2}

Simplify

- 9 \cdot \frac{ln ^{2} ( sin ( x ) )}{2} : - \frac{9}{2} ln^{2} (sin (x))

= - \frac{9}{2} ln^{2} (sin (x))

Add a constant to the solution

= - \frac{9}{2} ln^{2} (sin (x)) + C

\frac{d}{d x} (4 e^{3 x^{2}} sin (cos (2 x)))

\frac{d}{d x} (sin^{2} (x) - cos (x))

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

\frac{d}{d x} (7^{x} + 7 cos (\frac{7 x}{2}))

x^{2} y^{^{''}} - x y^{^{'}} + y = 0

What is the integral of-9cot(x)ln(sin(x)) ?
The integral of-9cot(x)ln(sin(x)) is -9/2 ln^2(sin(x))+C