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

The integral of sin^5(t)cos^4(t) is -(cos^5(t))/5+(2cos^7(t))/7-(cos^9(t))/9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^5(t)cos^4(t)

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

- \frac{cos ^{5} ( t )}{5} + \frac{2 cos ^{7} ( t )}{7} - \frac{cos ^{9} ( t )}{9} + C

Solution steps

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

Rewrite using trig identities

= \int (1 - cos^{2} (t))^{2} sin (t) cos^{4} (t) d t

Apply u-substitution

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

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

Expand

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

= \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

= - \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}

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

Substitute back

u = cos (t)

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

Add a constant to the solution

= - \frac{cos ^{5} ( t )}{5} + \frac{2 cos ^{7} ( t )}{7} - \frac{cos ^{9} ( t )}{9} + C

f^{^{'}} (x) = \frac{1 - x}{x ^{2}}

f (x) = \frac{2}{x + 4}, at x = - 3

\frac{d}{d x} (\frac{3}{x ^{2} - 2})

\int_{0}^{2 π} 2

\frac{20}{( s - 4 ) ( s ^{2} + 4 )}

What is the integral of sin^5(t)cos^4(t) ?
The integral of sin^5(t)cos^4(t) is -(cos^5(t))/5+(2cos^7(t))/7-(cos^9(t))/9+C