What is the integral of sin^5(8θ) ?

The integral of sin^5(8θ) is 1/8 (-cos(8θ)+(2cos^3(8θ))/3-(cos^5(8θ))/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(8θ)

\int sin^{5} (8 θ) d θ

\frac{1}{8} (- cos (8 θ) + \frac{2 cos ^{3} ( 8 θ )}{3} - \frac{cos ^{5} ( 8 θ )}{5}) + C

Solution steps

\int sin^{5} (8 θ) d θ

Apply u-substitution

: \int sin^{5} (u) \frac{1}{8} d u

= \int sin^{5} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

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

= \frac{1}{8} \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

= \frac{1}{8} (- \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}

= \frac{1}{8} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{8} (- cos (8 θ) + \frac{2 cos ^{3} ( 8 θ )}{3} - \frac{cos ^{5} ( 8 θ )}{5})

Add a constant to the solution

= \frac{1}{8} (- cos (8 θ) + \frac{2 cos ^{3} ( 8 θ )}{3} - \frac{cos ^{5} ( 8 θ )}{5}) + C

\frac{d}{d x} (arcsec (e^{x}))

(2) (1)

\int_{2}^{1} x^{2} d x

x \to 1 + lim (\frac{1}{x})

\frac{d}{d x} (e^{5 - x})

What is the integral of sin^5(8θ) ?
The integral of sin^5(8θ) is 1/8 (-cos(8θ)+(2cos^3(8θ))/3-(cos^5(8θ))/5)+C