What is the integral from 0 to pi/2 of cos^9(x)sin^5(x) ?

The integral from 0 to pi/2 of cos^9(x)sin^5(x) is 1/210

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of cos^9(x)sin^5(x)

\int_{0}^{\frac{π}{2}} cos^{9} (x) sin^{5} (x) d x

\frac{1}{210}

Decimal Notation

0.00476 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{9} (x) sin^{5} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} (1 - cos^{2} (x))^{2} sin (x) cos^{9} (x) d x

Apply u-substitution

: \int_{1}^{0} - u^{9} (1 - u^{2})^{2} d u

= \int_{1}^{0} - u^{9} (1 - u^{2})^{2} d u

Expand

- u^{9} (1 - u^{2})^{2} : - u^{9} + 2 u^{11} - u^{13}

= \int_{1}^{0} - u^{9} + 2 u^{11} - u^{13} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - u^{9} + 2 u^{11} - u^{13} 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

= - (- \int_{0}^{1} u^{9} d u + \int_{0}^{1} 2 u^{11} d u - \int_{0}^{1} u^{13} d u)

\int_{0}^{1} u^{9} d u = \frac{1}{10}

\int_{0}^{1} 2 u^{11} d u = \frac{1}{6}

\int_{0}^{1} u^{13} d u = \frac{1}{14}

= - (- \frac{1}{10} + \frac{1}{6} - \frac{1}{14})

Simplify

- (- \frac{1}{10} + \frac{1}{6} - \frac{1}{14}) : \frac{1}{210}

= \frac{1}{210}

\int_{- 2}^{3} \frac{3}{2 x ^{2}} d x

\int_{0}^{1} y^{2} 4 y^{2} + 1 d y

\int_{0}^{2} (x^{3} - 2 x) d x

\int_{0}^{\frac{π}{6}} tan^{3} (2 x) d x

\int_{0}^{x^{2}} \frac{1}{1 + t} d t

What is the integral from 0 to pi/2 of cos^9(x)sin^5(x) ?
The integral from 0 to pi/2 of cos^9(x)sin^5(x) is 1/210