What is the integral of 15sin^2(x)cos^3(x) ?

The integral of 15sin^2(x)cos^3(x) is 15((sin^3(x))/3-(sin^5(x))/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 15sin^2(x)cos^3(x)

\int 15 sin^{2} (x) cos^{3} (x) d x

15 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

Solution steps

\int 15 sin^{2} (x) cos^{3} (x) d x

Take the constant out:

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

= 15 \cdot \int sin^{2} (x) cos^{3} (x) d x

Rewrite using trig identities

= 15 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{2} (x) d x

Apply u-substitution

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

= 15 \cdot \int u^{2} (1 - u^{2}) d u

Expand

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

= 15 \cdot \int u^{2} - u^{4} 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

= 15 (\int u^{2} d u - \int u^{4} d u)

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

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

= 15 (\frac{u ^{3}}{3} - \frac{u ^{5}}{5})

Substitute back

u = sin (x)

= 15 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5})

Add a constant to the solution

= 15 (\frac{sin ^{3} ( x )}{3} - \frac{sin ^{5} ( x )}{5}) + C

\frac{d}{d x} (\frac{7}{2 x})

x \to - 1 lim (4 x^{3} + x^{2} + 2 x + 1)

\frac{d y}{d x} = y x^{9} sin (x^{10})

\frac{d}{d x} (\frac{1}{7 x})

\int (x^{2} - 1) (2 x + 1) d x

What is the integral of 15sin^2(x)cos^3(x) ?
The integral of 15sin^2(x)cos^3(x) is 15((sin^3(x))/3-(sin^5(x))/5)+C