What is the integral of sin^3(r)cos^6(r) ?

The integral of sin^3(r)cos^6(r) is -(cos^7(r))/7+(cos^9(r))/9+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^6(r)

\int sin^{3} (r) cos^{6} (r) d r

- \frac{cos ^{7} ( r )}{7} + \frac{cos ^{9} ( r )}{9} + C

Solution steps

\int sin^{3} (r) cos^{6} (r) d r

Rewrite using trig identities

= \int (1 - cos^{2} (r)) sin (r) cos^{6} (r) d r

Apply u-substitution

: \int - u^{6} (1 - u^{2}) d u

= \int - u^{6} (1 - u^{2}) d u

Expand

- u^{6} (1 - u^{2}) : - u^{6} + u^{8}

= \int - u^{6} + u^{8} 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^{6} d u + \int u^{8} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= - \frac{u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = cos (r)

= - \frac{cos ^{7} ( r )}{7} + \frac{cos ^{9} ( r )}{9}

Add a constant to the solution

= - \frac{cos ^{7} ( r )}{7} + \frac{cos ^{9} ( r )}{9} + C

\int e^{- 2 x + 18} d x

\int x^{2} x^{3} - 24 d x

\int (1 + 2 e^{t a n (y)}) sec^{2} (y) d y

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

\int + e^{- k x} d x

What is the integral of sin^3(r)cos^6(r) ?
The integral of sin^3(r)cos^6(r) is -(cos^7(r))/7+(cos^9(r))/9+C