What is the integral of sin^5(t)cos^6(t) ?

The integral of sin^5(t)cos^6(t) is -(cos^7(t))/7+(2cos^9(t))/9-(cos^{11}(t))/(11)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^5(t)cos^6(t)

\int sin^{5} (t) cos^{6} (t) d t

- \frac{cos ^{7} ( t )}{7} + \frac{2 cos ^{9} ( t )}{9} - \frac{cos ^{11} ( t )}{11} + C

Solution steps

\int sin^{5} (t) cos^{6} (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{6} (1 - u^{2})^{2} : - u^{6} + 2 u^{8} - u^{10}

= \int - u^{6} + 2 u^{8} - u^{10} 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 2 u^{8} d u - \int u^{10} d u

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

\int 2 u^{8} d u = \frac{2 u ^{9}}{9}

\int u^{10} d u = \frac{u ^{11}}{11}

= - \frac{u ^{7}}{7} + \frac{2 u ^{9}}{9} - \frac{u ^{11}}{11}

Substitute back

u = cos (t)

= - \frac{cos ^{7} ( t )}{7} + \frac{2 cos ^{9} ( t )}{9} - \frac{cos ^{11} ( t )}{11}

Add a constant to the solution

= - \frac{cos ^{7} ( t )}{7} + \frac{2 cos ^{9} ( t )}{9} - \frac{cos ^{11} ( t )}{11} + C

x \to 0 lim ((tan (9 x))^{x})

\frac{d}{d x} (- 4 π sin (8 π x))

\frac{d}{d x} (4 lo g_{3} (x^{2} - 1))

\frac{d}{d x} (20 x)

\int \frac{1}{x ( 1 + 4 I n ^{2} x )} d x

What is the integral of sin^5(t)cos^6(t) ?
The integral of sin^5(t)cos^6(t) is -(cos^7(t))/7+(2cos^9(t))/9-(cos^{11}(t))/(11)+C