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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (r)

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

Add a constant to the solution

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

\int 2^{- 3 x} d x

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

\int \frac{12 x ^{2} + 16 x + 3}{4 x ^{3} + 8 x ^{2} + 3 x} d x

\int x^{3} x - 3 d x

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

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