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

The integral of sin^4(5θ)cos^5(5θ) is 1/5 ((sin^5(5θ))/5-(2sin^7(5θ))/7+(sin^9(5θ))/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^4(5θ)cos^5(5θ)

\int sin^{4} (5 θ) cos^{5} (5 θ) d θ

\frac{1}{5} (\frac{sin ^{5} ( 5 θ )}{5} - \frac{2 sin ^{7} ( 5 θ )}{7} + \frac{sin ^{9} ( 5 θ )}{9}) + C

Solution steps

\int sin^{4} (5 θ) cos^{5} (5 θ) d θ

Apply u-substitution

: \int sin^{4} (u) cos^{5} (u) \frac{1}{5} d u

= \int sin^{4} (u) cos^{5} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sin^{4} (u) cos^{5} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 - sin^{2} (u))^{2} cos (u) sin^{4} (u) d u

Apply u-substitution

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

= \frac{1}{5} \cdot \int v^{4} (1 - v^{2})^{2} d v

Expand

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

= \frac{1}{5} \cdot \int v^{4} - 2 v^{6} + v^{8} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} (\int v^{4} d v - \int 2 v^{6} d v + \int v^{8} d v)

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

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

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

= \frac{1}{5} (\frac{v ^{5}}{5} - \frac{2 v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{1}{5} (\frac{sin ^{5} ( 5 θ )}{5} - \frac{2 sin ^{7} ( 5 θ )}{7} + \frac{sin ^{9} ( 5 θ )}{9})

Add a constant to the solution

= \frac{1}{5} (\frac{sin ^{5} ( 5 θ )}{5} - \frac{2 sin ^{7} ( 5 θ )}{7} + \frac{sin ^{9} ( 5 θ )}{9}) + C

f (x) = sin (x)

\frac{d y}{d x}, (x - 1)^{2} + (y + 4)^{2} = 25

y^{^{'}} - y = 2 x e^{2 x}, y (0) = 1

f (4)

\frac{8 x ^{2} + 2 x + 2}{x}

What is the integral of sin^4(5θ)cos^5(5θ) ?
The integral of sin^4(5θ)cos^5(5θ) is 1/5 ((sin^5(5θ))/5-(2sin^7(5θ))/7+(sin^9(5θ))/9)+C