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

The integral of sqrt(sin(x))cos^5(x) is 2/3 sin^{3/2}(x)-4/7 sin^{7/2}(x)+2/11 sin^{11/2}(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(sin(x))cos^5(x)

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

\frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{4}{7} sin^{\frac{7}{2}} (x) + \frac{2}{11} sin^{\frac{11}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u (1 - u^{2})^{2} : u - 2 u^{\frac{5}{2}} + u^{\frac{9}{2}}

= \int u - 2 u^{\frac{5}{2}} + u^{\frac{9}{2}} 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 u d u - \int 2 u^{\frac{5}{2}} d u + \int u^{\frac{9}{2}} d u

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

\int 2 u^{\frac{5}{2}} d u = \frac{4}{7} u^{\frac{7}{2}}

\int u^{\frac{9}{2}} d u = \frac{2}{11} u^{\frac{11}{2}}

= \frac{2}{3} u^{\frac{3}{2}} - \frac{4}{7} u^{\frac{7}{2}} + \frac{2}{11} u^{\frac{11}{2}}

Substitute back

u = sin (x)

= \frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{4}{7} sin^{\frac{7}{2}} (x) + \frac{2}{11} sin^{\frac{11}{2}} (x)

Add a constant to the solution

= \frac{2}{3} sin^{\frac{3}{2}} (x) - \frac{4}{7} sin^{\frac{7}{2}} (x) + \frac{2}{11} sin^{\frac{11}{2}} (x) + C

\frac{2}{s ^{2} ( s + 2 )}

\int 5 x + 3 d x

x^{3} - 3 x^{2} + 4, - 2, 3

\int \frac{x + 8}{x ^{2} + 4} d x

\int \frac{16 x}{8 x ^{2} + 1} d x

What is the integral of sqrt(sin(x))cos^5(x) ?
The integral of sqrt(sin(x))cos^5(x) is 2/3 sin^{3/2}(x)-4/7 sin^{7/2}(x)+2/11 sin^{11/2}(x)+C