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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^3(y)cos^4(y)

\int sin^{3} (y) cos^{4} (y) d y

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

Solution steps

\int sin^{3} (y) cos^{4} (y) d y

Rewrite using trig identities

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

Apply u-substitution

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

= \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 (y)

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

Add a constant to the solution

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

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

\int \frac{x + 7}{x ^{2} + 4 x + 6} d x

\int - x^{2} + 4 x + 2 d x

\int x x^{2} + 6 x + 13 d x

\int 5 x^{4} (x^{5} - 3)^{7} d x

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