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

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

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integral of sin(x)cos(x)ln(sin(x))

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

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

Solution steps

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

Apply u-substitution

: \int u ln (u) d u

= \int u ln (u) d u

Apply Integration By Parts

: \frac{1}{2} u^{2} ln (u) - \int \frac{u}{2} d u

= \frac{1}{2} u^{2} ln (u) - \int \frac{u}{2} d u

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

= \frac{1}{2} u^{2} ln (u) - \frac{u ^{2}}{4}

Substitute back

u = sin (x)

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

Add a constant to the solution

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{7}{3}

\int_{0}^{1} 4 - x^{2} d x

\int \frac{9}{x ^{4}} d x

y = \frac{x}{x + 1}, (4, \frac{2}{5})

x \to 0 lim (cos (x^{2}))

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