What is the integral of sin^2(x/8)cos(x/8) ?

The integral of sin^2(x/8)cos(x/8) is 8/3 sin^3(x/8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^2(x/8)cos(x/8)

\int sin^{2} (\frac{x}{8}) cos (\frac{x}{8}) d x

\frac{8}{3} sin^{3} (\frac{x}{8}) + C

Solution steps

\int sin^{2} (\frac{x}{8}) cos (\frac{x}{8}) d x

Apply u-substitution

: \int sin^{2} (u) cos (u) \cdot 8 d u

= \int sin^{2} (u) cos (u) \cdot 8 d u

Take the constant out:

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

= 8 \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= 8 \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= 8 \cdot \frac{v ^{3}}{3}

Substitute back

= 8 \cdot \frac{sin ^{3} ( \frac{x}{8} )}{3}

Simplify

8 \cdot \frac{sin ^{3} ( \frac{x}{8} )}{3} : \frac{8}{3} sin^{3} (\frac{x}{8})

= \frac{8}{3} sin^{3} (\frac{x}{8})

Add a constant to the solution

= \frac{8}{3} sin^{3} (\frac{x}{8}) + C

\int \frac{x}{49 - x ^{2}} d x

\int \frac{1 - x}{5 - 2 x - x ^{2}} d x

\int (3 x^{2} + 6 x + 5) d x

\int 2 z e^{- 2 yz} d y

\int ln (x) - x d x

What is the integral of sin^2(x/8)cos(x/8) ?
The integral of sin^2(x/8)cos(x/8) is 8/3 sin^3(x/8)+C