What is the integral from 0 to 1 of (1-x^2)^{5/2} ?

The integral from 0 to 1 of (1-x^2)^{5/2} is (5pi)/(32)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (1-x^2)^{5/2}

\int_{0}^{1} (1 - x^{2})^{\frac{5}{2}} d x

\frac{5 π}{32}

Decimal Notation

0.49087 \dots

Solution steps

\int_{0}^{1} (1 - x^{2})^{\frac{5}{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} cos^{6} (u) d u

= \int_{0}^{\frac{π}{2}} cos^{6} (u) d u

Apply Integral Reduction

= [\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} cos^{4} (u) d u

\int_{0}^{\frac{π}{2}} cos^{4} (u) d u = \frac{3 π}{16}

= [\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16}

Simplify

[\frac{sin ( u ) cos ^{5} ( u )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16} : [\frac{1}{6} cos^{5} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32}

= [\frac{1}{6} cos^{5} (u) sin (u)]_{0}^{\frac{π}{2}} + \frac{5 π}{32}

Compute the boundaries:

0

= 0 + \frac{5 π}{32}

Simplify

= \frac{5 π}{32}

\int_{- 2 π}^{3 π} cos^{2} (x) d x

\int_{1}^{3} \frac{1}{x ^{2} - 4 x + 5} d x

\int_{1}^{3} (\frac{9}{14} x^{2} - 2) d x

\int_{0}^{1 + s i n (x)} t^{2} d t

\int_{- 1}^{3} (7 - 2 x) d x

What is the integral from 0 to 1 of (1-x^2)^{5/2} ?
The integral from 0 to 1 of (1-x^2)^{5/2} is (5pi)/(32)