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

The integral of sin^3(x)cos^2(x is cos^2(d)(-xcos(x)+1/3 xcos^3(x)+2/3 sin(x)+1/9 sin^3(x))+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^2(x

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

cos^{2} (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

= cos^{2} (d) (x (- cos (x) + \frac{1}{3} cos^{3} (x)) - \int - cos (x) + \frac{1}{3} cos^{3} (x) d x)

\int - cos (x) + \frac{1}{3} cos^{3} (x) d x = - \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)

= cos^{2} (d) (x (- cos (x) + \frac{1}{3} cos^{3} (x)) - (- \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)))

Expand

x (- cos (x) + \frac{1}{3} cos^{3} (x)) - (- \frac{2}{3} sin (x) - \frac{1}{9} sin^{3} (x)) : - x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)

= cos^{2} (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x))

Add a constant to the solution

= cos^{2} (d) (- x cos (x) + \frac{1}{3} x cos^{3} (x) + \frac{2}{3} sin (x) + \frac{1}{9} sin^{3} (x)) + C

What is the integral of sin^3(x)cos^2(x ?
The integral of sin^3(x)cos^2(x is cos^2(d)(-xcos(x)+1/3 xcos^3(x)+2/3 sin(x)+1/9 sin^3(x))+C