What is the integral of sin(2x)ln(sin(x)) ?

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

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integral of sin(2x)ln(sin(x))

\int sin (2 x) ln (sin (x)) d x

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

Solution steps

\int sin (2 x) ln (sin (x)) d x

Rewrite using trig identities

= \int ln (sin (x)) \cdot 2 cos (x) sin (x) d x

Take the constant out:

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

= 2 \cdot \int ln (sin (x)) cos (x) sin (x) d x

Apply u-substitution

: \int u ln (u) d u

= 2 \cdot \int u ln (u) d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = sin (x)

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

Simplify

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

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

Add a constant to the solution

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

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