What is the integral of 14sin^2(x)cos^3(x) ?

The integral of 14sin^2(x)cos^3(x) is 14((sin^3(x))/3-(sin^5(x))/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 14sin^2(x)cos^3(x)

\int 14 sin^{2} (x) cos^{3} (x) d x

14 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

Solution steps

\int 14 sin^{2} (x) cos^{3} (x) d x

Take the constant out:

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

= 14 \cdot \int sin^{2} (x) cos^{3} (x) d x

Rewrite using trig identities

= 14 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{2} (x) d x

Apply u-substitution

: \int u^{2} (1 - u^{2}) d u

= 14 \cdot \int u^{2} (1 - u^{2}) d u

Expand

u^{2} (1 - u^{2}) : u^{2} - u^{4}

= 14 \cdot \int u^{2} - u^{4} d u

Apply the Sum Rule:

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

= 14 (\int u^{2} d u - \int u^{4} d u)

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

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

= 14 (\frac{u ^{3}}{3} - \frac{u ^{5}}{5})

Substitute back

u = sin (x)

= 14 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5})

Add a constant to the solution

= 14 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{c}{x} + k})

\frac{\partial}{\partial x} (- y (y + sin (x)))

1 - \frac{3 x}{x ^{3} + 27}

\int \frac{1}{x ^{2} + 12 x + 35} d x

(- 4.4) (- 1.5)

What is the integral of 14sin^2(x)cos^3(x) ?
The integral of 14sin^2(x)cos^3(x) is 14((sin^3(x))/3-(sin^5(x))/5)+C