What is the integral of ln(sin(x))cos^3(x) ?

The integral of ln(sin(x))cos^3(x) is -1/3 sin^3(x)ln(sin(x))+ln(sin(x))sin(x)+1/9 sin^3(x)-sin(x)+C

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integral of ln(sin(x))cos^3(x)

\int ln (sin (x)) cos^{3} (x) d x

- \frac{1}{3} sin^{3} (x) ln (sin (x)) + ln (sin (x)) sin (x) + \frac{1}{9} sin^{3} (x) - sin (x) + C

Solution steps

\int ln (sin (x)) cos^{3} (x) d x

Apply Trigonometric Substitution

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

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

Apply Integration By Parts

: ln (u) (- \frac{u ^{3}}{3} + u) - \int \frac{1}{3} (- u^{2} + 3) d u

= ln (u) (- \frac{u ^{3}}{3} + u) - \int \frac{1}{3} (- u^{2} + 3) d u

\int \frac{1}{3} (- u^{2} + 3) d u = - \frac{u ^{3}}{9} + u

= ln (u) (- \frac{u ^{3}}{3} + u) - (- \frac{u ^{3}}{9} + u)

Substitute back

u = sin (x)

= ln (sin (x)) (- \frac{sin ^{3} ( x )}{3} + sin (x)) - (- \frac{sin ^{3} ( x )}{9} + sin (x))

Simplify

ln (sin (x)) (- \frac{sin ^{3} ( x )}{3} + sin (x)) - (- \frac{sin ^{3} ( x )}{9} + sin (x)) : - \frac{1}{3} sin^{3} (x) ln (sin (x)) + ln (sin (x)) sin (x) + \frac{1}{9} sin^{3} (x) - sin (x)

= - \frac{1}{3} sin^{3} (x) ln (sin (x)) + ln (sin (x)) sin (x) + \frac{1}{9} sin^{3} (x) - sin (x)

Add a constant to the solution

= - \frac{1}{3} sin^{3} (x) ln (sin (x)) + ln (sin (x)) sin (x) + \frac{1}{9} sin^{3} (x) - sin (x) + C

t^{2} \frac{d y}{d t} + y = 0, y (5) = - 1, t > 0

x \to - 10 lim (\frac{1}{( x + 10 ) ^{4}})

\int 1 + 8 x d x

\int a r tan (x) d x

\int e^{- x} \frac{1}{x} d x

What is the integral of ln(sin(x))cos^3(x) ?
The integral of ln(sin(x))cos^3(x) is -1/3 sin^3(x)ln(sin(x))+ln(sin(x))sin(x)+1/9 sin^3(x)-sin(x)+C