What is the integral of sin^5(x/2) ?

The integral of sin^5(x/2) is 2(-cos(x/2)+(2cos^3(x/2))/3-(cos^5(x/2))/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^5(x/2)

\int sin^{5} (\frac{x}{2}) d x

2 (- cos (\frac{x}{2}) + \frac{2 cos ^{3} ( \frac{x}{2} )}{3} - \frac{cos ^{5} ( \frac{x}{2} )}{5}) + C

Solution steps

\int sin^{5} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{5} (u) \cdot 2 d u

= \int sin^{5} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= 2 \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

: \int - 1 + 2 v^{2} - v^{4} d v

= 2 \cdot \int - 1 + 2 v^{2} - v^{4} 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

= 2 (- \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= 2 (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= 2 (- cos (\frac{x}{2}) + \frac{2 cos ^{3} ( \frac{x}{2} )}{3} - \frac{cos ^{5} ( \frac{x}{2} )}{5})

Add a constant to the solution

= 2 (- cos (\frac{x}{2}) + \frac{2 cos ^{3} ( \frac{x}{2} )}{3} - \frac{cos ^{5} ( \frac{x}{2} )}{5}) + C

\frac{d}{d x} (x^{2} sin (ln (x)))

x \to 0 lim (5 - 2 x^{2})

x \to 3 lim (2 x^{- 1})

\frac{d y}{d x} = \frac{1}{4} y cos^{2} (y)

\int csc (x) tan (x) d x

What is the integral of sin^5(x/2) ?
The integral of sin^5(x/2) is 2(-cos(x/2)+(2cos^3(x/2))/3-(cos^5(x/2))/5)+C