What is the integral of sin^5(11x) ?

The integral of sin^5(11x) is 1/11 (-cos(11x)+(2cos^3(11x))/3-(cos^5(11x))/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(11x)

\int sin^{5} (11 x) d x

\frac{1}{11} (- cos (11 x) + \frac{2 cos ^{3} ( 11 x )}{3} - \frac{cos ^{5} ( 11 x )}{5}) + C

Solution steps

\int sin^{5} (11 x) d x

Apply u-substitution

: \int sin^{5} (u) \frac{1}{11} d u

= \int sin^{5} (u) \frac{1}{11} d u

Take the constant out:

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

= \frac{1}{11} \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{11} \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}{11} \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}{11} (- \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}{11} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{11} (- cos (11 x) + \frac{2 cos ^{3} ( 11 x )}{3} - \frac{cos ^{5} ( 11 x )}{5})

Add a constant to the solution

= \frac{1}{11} (- cos (11 x) + \frac{2 cos ^{3} ( 11 x )}{3} - \frac{cos ^{5} ( 11 x )}{5}) + C

\int ([sin (x)]^{2} x + [cos (x)]^{2} x) d x

\int \frac{( 1 + x ) ^{3}}{x} d x

\int_{1}^{\infty} \frac{1}{4 + x ^{2}} d x

\frac{d}{d x} ((2 x - 1)^{3} (3 x + 2)^{4})

\frac{d y}{d x} = tan (x)

What is the integral of sin^5(11x) ?
The integral of sin^5(11x) is 1/11 (-cos(11x)+(2cos^3(11x))/3-(cos^5(11x))/5)+C