What is the integral of sin^3(x)cos^5(x) ?

The integral of sin^3(x)cos^5(x) is -(cos^6(x))/6+(cos^8(x))/8+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(x)cos^5(x)

\int sin^{3} (x) cos^{5} (x) d x

- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8} + C

Solution steps

\int sin^{3} (x) cos^{5} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

: \int - u^{5} (1 - u^{2}) d u

= \int - u^{5} (1 - u^{2}) d u

Expand

- u^{5} (1 - u^{2}) : - u^{5} + u^{7}

= \int - u^{5} + u^{7} 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^{5} d u + \int u^{7} d u

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

\int u^{7} d u = \frac{u ^{8}}{8}

= - \frac{u ^{6}}{6} + \frac{u ^{8}}{8}

Substitute back

u = cos (x)

= - \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}

Add a constant to the solution

= - \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8} + C

\frac{d}{d x} (\frac{2 x}{1 - 2 x})

\int arcsin (6 x) d x

(3, 1), (0, - 2)

5 \cdot 2^{t - 2}

n = 4 \sum \infty e^{7 - 4 n}

What is the integral of sin^3(x)cos^5(x) ?
The integral of sin^3(x)cos^5(x) is -(cos^6(x))/6+(cos^8(x))/8+C