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

The integral of cos^3(y)sin^3(y) is -(cos^4(y))/4+(cos^6(y))/6+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of cos^3(y)sin^3(y)

\int cos^{3} (y) sin^{3} (y) d y

- \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{6} + C

Solution steps

\int cos^{3} (y) sin^{3} (y) d y

Rewrite using trig identities

= \int (1 - cos^{2} (y)) sin (y) cos^{3} (y) d y

Apply u-substitution

= \int - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= \int - u^{3} + u^{5} 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^{3} d u + \int u^{5} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = cos (y)

= - \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{6}

Add a constant to the solution

= - \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{6} + C

\int 2 - 2 tan^{2} (x) d x

\int \frac{2 y ^{4}}{y ^{5} + 1} d y

\int (\frac{5}{3} x^{3} x^{4} - 3) d x

\int \frac{x ^{2} - 4}{x ^{3} - 12 x + 9} d x

\int \frac{1}{x ^{4} + 3 x ^{2}} d x

What is the integral of cos^3(y)sin^3(y) ?
The integral of cos^3(y)sin^3(y) is -(cos^4(y))/4+(cos^6(y))/6+C