What is the integral from ln(x) to 1 of sin^3(u) ?

The integral from ln(x) to 1 of sin^3(u) is -cos(1)+cos(ln(x))+(cos^3(1)-cos^3(ln(x)))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from ln(x) to 1 of sin^3(u)

\int_{l n (x)}^{1} sin^{3} (u) d u

- cos (1) + cos (ln (x)) + \frac{cos ^{3} ( 1 ) - cos ^{3} ( ln ( x ) )}{3}

Solution steps

\int_{l n (x)}^{1} sin^{3} (u) d u

Rewrite using trig identities

= \int_{l n (x)}^{1} (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

: \int_{c o s (l n (x))}^{c o s (1)} - 1 + v^{2} d v

= \int_{c o s (l n (x))}^{c o s (1)} - 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{c o s (l n (x))}^{c o s (1)} 1 d v + \int_{c o s (l n (x))}^{c o s (1)} v^{2} d v

\int_{c o s (l n (x))}^{c o s (1)} 1 d v = cos (1) - cos (ln (x))

\int_{c o s (l n (x))}^{c o s (1)} v^{2} d v = \frac{cos ^{3} ( 1 ) - cos ^{3} ( ln ( x ) )}{3}

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

Simplify

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

y = x^{3} - 12 x^{2} + 32 x, - x^{3} + 12 x^{2} - 32 x

\int e^{\frac{x}{y}} d x

x \to 2 lim (sin (\frac{2}{x - 2}))

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

\int - \frac{1}{2} cos (2 x) + C d x

What is the integral from ln(x) to 1 of sin^3(u) ?
The integral from ln(x) to 1 of sin^3(u) is -cos(1)+cos(ln(x))+(cos^3(1)-cos^3(ln(x)))/3