What is the integral of (cot(x))[ln(sin(x))] ?

The integral of (cot(x))[ln(sin(x))] is (ln^2(sin(x)))/2+C

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integral of (cot(x))[ln(sin(x))]

\int (cot (x)) [ln (sin (x))] d x

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

Solution steps

\int cot (x) [ln (sin (x))] d x

Simplify

= \int cot (x) ln (sin (x)) d x

Rewrite using trig identities

= \int cos (x) ln (sin (x)) csc (x) d x

Apply u-substitution

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

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

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

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

= \frac{v ^{2}}{2}

Substitute back

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

Add a constant to the solution

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

\int x^{2} + x + 6 d x

\int \frac{3 x ^{2} + 5 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

(x + 2 y^{2}) d x + x y d y = 0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x, y (1) = 0, y^{^{'}} (1) = 0

\frac{d y}{d x} = \frac{x}{x ^{2} + 64}

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