What is the integral of 9cos^5(x)sin^4(x) ?

The integral of 9cos^5(x)sin^4(x) is 9/5 sin^5(x)-18/7 sin^7(x)+sin^9(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9cos^5(x)sin^4(x)

\int 9 cos^{5} (x) sin^{4} (x) d x

\frac{9}{5} sin^{5} (x) - \frac{18}{7} sin^{7} (x) + sin^{9} (x) + C

Solution steps

\int 9 cos^{5} (x) sin^{4} (x) d x

Take the constant out:

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

= 9 \cdot \int cos^{5} (x) sin^{4} (x) d x

Rewrite using trig identities

= 9 \cdot \int (1 - sin^{2} (x))^{2} cos (x) sin^{4} (x) d x

Apply u-substitution

: \int u^{4} (1 - u^{2})^{2} d u

= 9 \cdot \int u^{4} (1 - u^{2})^{2} d u

Expand

u^{4} (1 - u^{2})^{2} : u^{4} - 2 u^{6} + u^{8}

= 9 \cdot \int u^{4} - 2 u^{6} + u^{8} 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

= 9 (\int u^{4} d u - \int 2 u^{6} d u + \int u^{8} d u)

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 9 (\frac{u ^{5}}{5} - \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = sin (x)

= 9 (\frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9})

Simplify

9 (\frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9}) : \frac{9}{5} sin^{5} (x) - \frac{18}{7} sin^{7} (x) + sin^{9} (x)

= \frac{9}{5} sin^{5} (x) - \frac{18}{7} sin^{7} (x) + sin^{9} (x)

Add a constant to the solution

= \frac{9}{5} sin^{5} (x) - \frac{18}{7} sin^{7} (x) + sin^{9} (x) + C

\frac{\partial}{\partial x} (2 x y + 3 x - 4 y + 1)

\frac{\partial}{\partial x} (x y^{7} - x^{2} y)

\frac{d}{d x} (10 e^{2 x})

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

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

What is the integral of 9cos^5(x)sin^4(x) ?
The integral of 9cos^5(x)sin^4(x) is 9/5 sin^5(x)-18/7 sin^7(x)+sin^9(x)+C