What is the integral of sin^3(θ)cos^4(θ) ?

The integral of sin^3(θ)cos^4(θ) is -(cos^5(θ))/5+(cos^7(θ))/7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin^3(θ)cos^4(θ)

\int sin^{3} (θ) cos^{4} (θ) d θ

- \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7} + C

Solution steps

\int sin^{3} (θ) cos^{4} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

Expand

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

= \int - u^{4} + u^{6} 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 u^{6} d u

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

\int u^{6} d u = \frac{u ^{7}}{7}

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

Substitute back

u = cos (θ)

= - \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7}

Add a constant to the solution

= - \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7} + C

\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}

\frac{d}{d x} ((2 x^{3} - 5) (x^{2} - 5 ln (x)))

\frac{d}{d x} (\frac{x}{csc ( x )})

\frac{\partial}{\partial z} (x^{2} - x yz)

y^{^{''}} - y^{^{'}} - 42 y = 0

What is the integral of sin^3(θ)cos^4(θ) ?
The integral of sin^3(θ)cos^4(θ) is -(cos^5(θ))/5+(cos^7(θ))/7+C