What is the integral of sin^3(x)cos^{12}(x) ?

The integral of sin^3(x)cos^{12}(x) is -(cos^{13}(x))/(13)+(cos^{15}(x))/(15)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin^3(x)cos^{12}(x)

\int sin^{3} (x) cos^{12} (x) d x

- \frac{cos ^{13} ( x )}{13} + \frac{cos ^{15} ( x )}{15} + C

Solution steps

\int sin^{3} (x) cos^{12} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{12} (x) d x

Apply u-substitution

= \int - u^{12} (1 - u^{2}) d u

Expand

- u^{12} (1 - u^{2}) : - u^{12} + u^{14}

= \int - u^{12} + u^{14} 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^{12} d u + \int u^{14} d u

\int u^{12} d u = \frac{u ^{13}}{13}

\int u^{14} d u = \frac{u ^{15}}{15}

= - \frac{u ^{13}}{13} + \frac{u ^{15}}{15}

Substitute back

u = cos (x)

= - \frac{cos ^{13} ( x )}{13} + \frac{cos ^{15} ( x )}{15}

Add a constant to the solution

= - \frac{cos ^{13} ( x )}{13} + \frac{cos ^{15} ( x )}{15} + C

\int_{0}^{\infty} \frac{1}{x ^{4} - 1} d x

\int_{7}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{\partial}{\partial y} (y cos (6 x))

t an g e n t f (x) = x at (25.5)

\int_{1}^{3} (u - 2)^{4} d u

What is the integral of sin^3(x)cos^{12}(x) ?
The integral of sin^3(x)cos^{12}(x) is -(cos^{13}(x))/(13)+(cos^{15}(x))/(15)+C